Varieties of elements of given order in simple algebraic groups

TL;DR

This paper investigates the structure and properties of elements of specific orders in simple algebraic groups, determining the dimension of related subvarieties and applying findings to finite simple quotients of triangle groups.

Contribution

It provides a complete determination of the dimension of subvarieties of elements of given order in simple algebraic groups, extending previous results and applying to finite group quotients.

Findings

Determined the dimension of subvarieties of elements of order dividing u in simple algebraic groups.

Extended Lawther's results to groups of adjoint type.

Proved certain finite quasisimple groups are not quotients of specific triangle groups.

Abstract

Given a positive integer $u$ and a simple algebraic group $G$ defined over an algebraically closed field $K$ of characteristic $p$, we derive properties about the subvariety $G_{[u]}$ of $G$ consisting of elements of $G$ of order dividing $u$. In particular, we determine the dimension of $G_{[u]}$, completing results of Lawther [7] in the special case where $G$ is of adjoint type. We also apply our results to the study of finite simple quotients of triangle groups, giving further insight on a conjecture we proposed in [10] as well as proving that some finite quasisimple groups are not quotients of certain triangle groups.

Tables

Table 1. Table 1: Reducible triples for simple algebraic groups of simply connected or adjoint type

$G$	$p$	$(a,b,c)$
${\mathrm{SL}}_2\left(K\right)$	$p\ne 2$	$(2,b,c)$
${\mathrm{Sp}}_4\left(K\right)$	$p\ne 2$	$(2,3,c)$, $(2,4,c)$, $(3,3,4)$, $(3,4,4)$, $(4,4,4)$
${\mathrm{Sp}}_6\left(K\right)$	$p\ne 2$	$(2,3,c)$, $(2,4,c)$, $(2,5,5)$, $(2,5,6)$, $(2,6,6)$
${\mathrm{Sp}}_8\left(K\right)$	$p\ne 2$	$(2,3,7)$, $(2,3,8)$, $(2,4,5)$, $(2,4,6)$
${\mathrm{Sp}}_{10}\left(K\right)$	$p\ne 2$	$(2,3,7)$, $(2,3,8)$, $(2,3,9)$, $(2,3,10)$, $(2,4,5)$, $(2,4,6)$
${\mathrm{Sp}}_{12}\left(K\right)$	$p\ne 2$	$(2,3,7)$, $(2,3,8)$, $(2,4,5)$
${\mathrm{Sp}}_{14}\left(K\right)$	$p\ne 2$	$(2,3,7)$, $(2,3,8)$, $(2,4,5)$
${\mathrm{Sp}}_{16}\left(K\right)$	$p\ne 2$	$(2,3,7)$
${\mathrm{Sp}}_{18}\left(K\right)$	$p\ne 2$	$(2,3,7)$
${\mathrm{Sp}}_{22}\left(K\right)$	$p\ne 2$	$(2,3,7)$

Table 2. Table 2: Rigid triples for simple algebraic groups of simply connected type

$G$	$p$	$(a,b,c)$
${\mathrm{SL}}_2\left(K\right)$	$p=2$	$(a,b,c)$
	$p\ne 2$	$(a,b,c)$ $a\ge 3$
${\mathrm{SL}}_3\left(K\right)$	any	$(2,b,c)$
${\mathrm{SL}}_4\left(K\right)$	$p=2$	$(2,3,c)$
	$p\ne 2$	$(2,3,c)$, $(2,4,c)$, $(3,3,4)$, $(3,4,4)$,$(4,4,4)$
${\mathrm{SL}}_5\left(K\right)$	any	$(2,3,c)$
${\mathrm{SL}}_6\left(K\right)$	$p\ne 2$	$(2,3,c)$, $(2,4,5)$, $(2,4,6)$
${\mathrm{SL}}_{10}\left(K\right)$	$p\ne 2$	$(2,3,7)$
${\mathrm{Sp}}_4\left(K\right)$	$p=2$	$(2,3,c)$, $(3,3,c)$
${\mathrm{Sp}}_4\left(K\right)$	$p\ne 2$	$(2,b,c)$ $b\ge 5$, $(3,3,c)$ $c\ge 5$, $(3,4,c)$ $c\ge 5$, $(4,4,c)$ $c\ge 5$
${\mathrm{Sp}}_6\left(K\right)$	$p\ne 2$	$(2,5,c)$ $c\ge 7$, $(2,6,c)$ $c\ge 7$, $(3,3,4)$, $(3,4,4)$, $(4,4,4)$
${\mathrm{Sp}}_8\left(K\right)$	$p\ne 2$	$(2,3,c)$ $c\ge 9$, $(2,4,7)$, $(2,4,8)$, $(2,5,5)$, $(2,5,6)$, $(2,6,6)$
${\mathrm{Sp}}_{10}\left(K\right)$	$p\ne 2$	$(2,3,c)$ $c\ge 11$, $(2,4,7)$, $(2,4,8)$
${\mathrm{Sp}}_{12}\left(K\right)$	$p\ne 2$	$(2,3,9)$, $(2,3,10)$, $(2,4,6)$
${\mathrm{Sp}}_{14}\left(K\right)$	$p\ne 2$	$(2,3,9)$, $(2,3,10)$, $(2,4,6)$
${\mathrm{Sp}}_{16}\left(K\right)$	$p\ne 2$	$(2,3,8)$, $(2,4,5)$
${\mathrm{Sp}}_{18}\left(K\right)$	$p\ne 2$	$(2,3,8)$, $(2,4,5)$
${\mathrm{Sp}}_{20}\left(K\right)$	$p\ne 2$	$(2,3,7)$
${\mathrm{Sp}}_{24}\left(K\right)$	$p\ne 2$	$(2,3,7)$
${\mathrm{Sp}}_{26}\left(K\right)$	$p\ne 2$	$(2,3,7)$
${\mathrm{Spin}}_{11}\left(K\right)$	$p\ne 2$	$(2,3,7)$
${\mathrm{Spin}}_{12}\left(K\right)$	$p\ne 2$	$(2,3,7)$

Table 3. Table 3: Rigid triples for simple algebraic groups of adjoint type

$G$	$p$	$(a,b,c)$
${\mathrm{PSL}}_2\left(K\right)$	any	$(a,b,c)$
${\mathrm{PSL}}_3\left(K\right)$	any	$(2,b,c)$
${\mathrm{PSL}}_4\left(K\right)$	any	$(2,3,c)$
${\mathrm{PSL}}_5\left(K\right)$	any	$(2,3,c)$
${\mathrm{PSp}}_4\left(K\right)$	any	$(2,3,c)$, $(3,3,c)$
$G_2\left(K\right)$	any	$(2,4,5)$, (2,5,5)

Table 4. Table 4: A semisimple element y 𝑦 y of G = ( A ℓ ) s . c 𝐺 subscript subscript 𝐴 ℓ formulae-sequence s c G=(A_{\ell})_{\rm{s.c}} of order v 𝑣 v giving an upper bound for d u ​ ( G ) subscript 𝑑 𝑢 𝐺 d_{u}(G)

Case	Element $y$	$C_{G_{s.c.}}{\left(y\right)}^0$	$d_q\left(C_{G_{s.c.}}{\left(y\right)}^0\right)$
$\begin{array}{c}{ϵ}_v=1,\\ \text{or}({ϵ}_v,{ϵ}_{\alpha })=(0,0)\end{array}$	$M_1^{\alpha }\oplus M_2\oplus {\left(1\right)}^{{ϵ}_{\beta }}$	$A_{\alpha }^{\beta }{A}_{\alpha -1}^{v-\beta }{T}_{v-1}$	$z^{2}u+e\left(2z+1\right)-1$
$({ϵ}_v,{ϵ}_{\alpha },{ϵ}_{\beta })=(0,1,1)$	$M_1^{\alpha }\oplus M_2\oplus \left(-1\right)$
$\begin{array}{c}({ϵ}_v,{ϵ}_{\alpha },{ϵ}_{\beta })=(0,1,0),\\ \beta \ge 2\end{array}$	$M_1^{\alpha }\oplus M_5\oplus \left(1\right)\oplus \left(-1\right)$
$({ϵ}_v,{ϵ}_{\alpha })=(0,1)$, $\beta =0$	$M_1^{\alpha -1}\oplus M_4\oplus \left(-1\right)$	$A_{\alpha }{A}_{\alpha -1}^{v-2}{A}_{\alpha -2}{T}_{v-1}$

Table 5. Table 5: A semisimple element y 𝑦 y of G = ( C ℓ ) s . c 𝐺 subscript subscript 𝐶 ℓ formulae-sequence s c G=(C_{\ell})_{\rm{s.c}} of order v 𝑣 v giving an upper bound for d u ​ ( G ) subscript 𝑑 𝑢 𝐺 d_{u}(G)

Case	Element $y$	$C_G{\left(y\right)}^0$	$d_q\left(C_G{\left(y\right)}^0\right)$
${ϵ}_v=1$	$M_1^{\alpha }\oplus M_2\oplus {\left(1\right)}^{{ϵ}_{\beta }}$	$A_{\alpha }^{\lfloor \frac{\beta }{2}\rfloor }{A}_{\alpha -1}^{\frac{v-1}{2}-\lfloor \frac{\beta }{2}\rfloor }{C}_{\lceil \frac{\alpha }{2}\rceil }{T}_{\frac{v-1}{2}}$	$\frac{\left(z^{2}u+e\left(2z+1\right)\right)}{2}+\lceil \frac{z}{2}\rceil {ϵ}_u$
$({ϵ}_v,{ϵ}_{\alpha })=(0,0)$	$M_1^{\alpha }\oplus M_2$	$A_{\alpha }^{\frac{\beta }{2}}{A}_{\alpha -1}^{\frac{v}{2}-1-\frac{\beta }{2}}{C}_{\frac{\alpha }{2}}^2{T}_{\frac{v}{2}-1}$	$\frac{\left(z^{2}u+e\left(2z+1\right)\right)}{2}+2\lceil \frac{z}{2}\rceil$
$({ϵ}_v,{ϵ}_{\alpha })=(0,1)$, $\beta \ge 2$	$M_1^{\alpha }\oplus M_5\oplus \left(1\right)\oplus \left(-1\right)$	$A_{\alpha }^{\frac{\beta }{2}-1}{A}_{\alpha -1}^{\frac{v}{2}-\frac{\beta }{2}}{C}_{\frac{\alpha +1}{2}}^2{T}_{\frac{v}{2}-1}$
$({ϵ}_v,{ϵ}_{\alpha })=(0,1)$, $\beta =0$	$M_1^{\alpha -1}\oplus M_4\oplus \left(-1\right)$	$A_{\alpha -1}^{\frac{v}{2}-1}{C}_{\frac{\alpha +1}{2}}{C}_{\frac{\alpha -1}{2}}{T}_{\frac{v}{2}-1}$

Table 6. Table 6: A semisimple element y 𝑦 y of G = ( B ℓ ) s . c 𝐺 subscript subscript 𝐵 ℓ formulae-sequence s c G=(B_{\ell})_{\rm{s.c}} of order v 𝑣 v giving an upper bound for d u ​ ( G ) subscript 𝑑 𝑢 𝐺 d_{u}(G)

Case	Element $\overline{y}$	$C_{{\mathrm{SO}}_{h+1}\left(K\right)}{\left(\overline{y}\right)}^0$	$d_q\left(C_G{\left(y\right)}^0\right)$
(a)	$M_1^{\alpha }\oplus M_2\oplus {\left(1\right)}^{1+{ϵ}_{\beta }}$	$A_{\alpha }^{\lfloor \frac{\beta }{2}\rfloor }{A}_{\alpha -1}^{\frac{v-1}{2}-\lfloor \frac{\beta }{2}\rfloor }{B}_{\lceil \frac{\alpha }{2}\rceil }{T}_{\frac{v-1}{2}}$	$\frac{1}{2}\left(z^{2}u+e\left(2z+1\right)\right)+\lceil \frac{z}{2}\rceil {ϵ}_u$
(b)
(c)	$M_1^{\alpha }\oplus M_2\oplus \left(-1\right)$	$A_{\alpha }^{\frac{\beta }{2}}{A}_{\alpha -1}^{\frac{v}{2}-1-\frac{\beta }{2}}{B}_{\frac{\alpha -1}{2}}{D}_{\frac{\alpha +1}{2}}{T}_{\frac{v}{2}-1}$	$\frac{1}{2}\left(z^{2}u+e\left(2z+1\right)\right)$
(d)	$M_1^{\alpha }\oplus M_5\oplus {\left(1\right)}^2\oplus \left(-1\right)$	$A_{\alpha }^{\frac{\beta }{2}-1}{A}_{\alpha -1}^{\frac{v}{2}-\frac{\beta }{2}}{B}_{\frac{\alpha +1}{2}}{D}_{\frac{\alpha +1}{2}}{T}_{\frac{v}{2}-1}$
(e)	$M_1^{\alpha }\oplus M_6\oplus {\left(1\right)}^2\oplus \left(-1\right)$
(f)	$M_1^{\alpha -1}\oplus M_7\oplus {\left(1\right)}^3\oplus {\left(-1\right)}^2$	$A_{\alpha -1}^{\frac{v}{2}-2}{A}_{\alpha -2}{B}_{\frac{\alpha +1}{2}}{D}_{\frac{\alpha +1}{2}}{T}_{\frac{v}{2}-1}$
(g)	$M_1^{\alpha }\oplus M_2\oplus \left(1\right)$	$A_{\alpha }^{\frac{\beta }{2}}{A}_{\alpha -1}^{\frac{v}{2}-1-\frac{\beta }{2}}{B}_{\frac{\alpha }{2}}{D}_{\frac{\alpha }{2}}{T}_{\frac{v}{2}-1}$	$\frac{1}{2}\left(z^{2}u+e\left(2z+1\right)\right)$
(h)	$M_1^{\alpha }\oplus M_6\oplus \left({\omega }^{\frac{v}{2}-1}\right)\oplus \left({\omega }^{-\left(\frac{v}{2}-1\right)}\right)\oplus \left(1\right)$
(i)	$M_1^{\alpha }\oplus M_5\oplus \left(1\right)\oplus {\left(-1\right)}^2$	$A_{\alpha }^{\frac{\beta }{2}-1}{A}_{\alpha -1}^{\frac{v}{2}-\frac{\beta }{2}}{B}_{\frac{\alpha }{2}}{D}_{\frac{\alpha +2}{2}}{T}_{\frac{v}{2}-1}$
(j)	$M_1^{\alpha }\oplus M_6\oplus \left(1\right)\oplus {\left(-1\right)}^2$
(k)	$M_1^{\alpha -1}\oplus M_9\oplus {\left(1\right)}^2\oplus {\left(-1\right)}^3$	$A_{\alpha -1}^{\frac{v}{2}-2}{A}_{\alpha -2}{B}_{\frac{\alpha }{2}}{D}_{\frac{\alpha +2}{2}}{T}_{\frac{v}{2}-1}$

Table 7. Table 7: A semisimple element y 𝑦 y of G = ( D ℓ ) s . c 𝐺 subscript subscript 𝐷 ℓ formulae-sequence s c G=(D_{\ell})_{\rm{s.c}} of order v 𝑣 v giving an upper bound for d u ​ ( G ) subscript 𝑑 𝑢 𝐺 d_{u}(G)

Case	Element $\overline{y}$	$C_{{\mathrm{SO}}_{h+1}\left(K\right)}{\left(\overline{y}\right)}^0$	$d_q\left(C_G{\left(y\right)}^0\right)$
(a)	$M_1^{\alpha }\oplus M_3\oplus {\left(1\right)}^{2-{ϵ}_{\beta }}$	$A_{\alpha }^{\lceil \frac{\beta }{2}\rceil }{A}_{\alpha -1}^{\frac{v-1}{2}-\lceil \frac{\beta }{2}\rceil }{D}_{\lceil \frac{\alpha +1}{2}\rceil }{T}_{\frac{v-1}{2}}$	$\frac{1}{2}\left(z^{2}u+e\left(2z+1\right)\right)+\lceil \frac{z}{2}\rceil {ϵ}_u+z+1-{ϵ}_z$
(b)
(c)	$M_1^{\alpha }\oplus M_3\oplus \left(1\right)\oplus \left(-1\right)$	$A_{\alpha }^{\frac{\beta }{2}}{A}_{\alpha -1}^{\frac{v}{2}-1-\frac{\beta }{2}}{D}_{\frac{\alpha +1}{2}}{D}_{\frac{\alpha +1}{2}}{T}_{\frac{v}{2}-1}$	$\frac{z^{2}u+e\left(2z+1\right)}{2}+z+1-{ϵ}_z$
(d)	$M_1^{\alpha }\oplus M_5\oplus \left({\omega }^{\frac{v}{2}-2}\right)\oplus \left({\omega }^{-\left(\frac{v}{2}-2\right)}\right)\oplus \left(1\right)\oplus \left(-1\right)$
(e)	$M_1^{\alpha }\oplus M_5\oplus \left({\omega }^{\frac{v}{2}-1}\right)\oplus \left({\omega }^{-\left(\frac{v}{2}-1\right)}\right)\oplus \left(1\right)\oplus \left(-1\right)$
(f)	$M_1^{\alpha }\oplus M_6\oplus \left({\omega }^{\frac{v}{2}-2}\right)\oplus \left({\omega }^{-\left(\frac{v}{2}-2\right)}\right)\oplus \left(1\right)\oplus \left(-1\right)$
	if $\beta \ne v-4$
	$M_1^{\alpha }\oplus M_8\oplus \left(1\right)\oplus \left(-1\right)$
	if $\beta =v-4$
(g)	$M_1^{\alpha }\oplus M_6\oplus \left({\omega }^{\frac{v}{2}-1}\right)\oplus \left({\omega }^{-\left(\frac{v}{2}-1\right)}\right)\oplus \left(1\right)\oplus \left(-1\right)$
(h)	$M_1^{\alpha }\oplus M_9\oplus \left(1\right)\oplus {\left(-1\right)}^3$	$A_{\alpha }^{\frac{v}{2}-2}{A}_{\alpha -1}{D}_{\frac{\alpha +1}{2}}{D}_{\frac{\alpha +3}{2}}{T}_{\frac{v}{2}-1}$
(i)	$M_1^{\alpha -1}\oplus M_7\oplus {\left(\omega \right)}^2\oplus {\left({\omega }^{-1}\right)}^2\oplus {\left(1\right)}^2$	$A_{\alpha }{A}_{\alpha -1}^{\frac{v}{2}-2}{D}_{\frac{\alpha -1}{2}}{D}_{\frac{\alpha +1}{2}}{T}_{\frac{v}{2}-1}$
(j)	$M_1^{\alpha -1}\oplus M_9\oplus \left(\omega \right)\oplus \left({\omega }^{-1}\right)\oplus {\left(1\right)}^2\oplus {\left(-1\right)}^2$	$A_{\alpha }{A}_{\alpha -1}^{\frac{v}{2}-3}{A}_{\alpha -2}{D}_{\frac{\alpha +1}{2}}^2{T}_{\frac{v}{2}-1}$
(k)	$M_1^{\alpha }\oplus M_3\oplus {\left(1\right)}^2$	$A_{\alpha }^{\frac{\beta }{2}}{A}_{\alpha -1}^{\frac{v}{2}-1-\frac{\beta }{2}}{D}_{\frac{\alpha }{2}}{D}_{\frac{\alpha +2}{2}}{T}_{\frac{v}{2}-1}$	$\frac{z^{2}u+e\left(2z+1\right)}{2}+z+1-{ϵ}_z$
(l)	$M_1^{\alpha }\oplus M_5\oplus \left({\omega }^{\frac{v}{2}-1}\right)\oplus \left({\omega }^{-\left(\frac{v}{2}-1\right)}\right)\oplus {\left(1\right)}^2$
(m)	$M_1^{\alpha }\oplus M_6\oplus \left({\omega }^{\frac{v}{2}-2}\right)\oplus \left({\omega }^{-\left(\frac{v}{2}-2\right)}\right)\oplus {\left(1\right)}^2$
	if $\beta \ne v-4$
	$M_1^{\alpha }\oplus M_8\oplus {\left(1\right)}^2$
	if $\beta =v-4$
(n)	$M_1^{\alpha }\oplus M_5\oplus \left({\omega }^{\frac{v}{2}-2}\right)\oplus \left({\omega }^{-\left(\frac{v}{2}-2\right)}\right)\oplus {\left(1\right)}^2$	$A_{\alpha }^{\frac{\beta }{2}}{A}_{\alpha -1}^{\frac{v}{2}-1-\frac{\beta }{2}}{D}_{\frac{\alpha }{2}}{D}_{\frac{\alpha +2}{2}}{T}_{\frac{v}{2}-1}$
	if $\beta \ne v-2$	if $\beta \ne v-2$
	$M_1^{\alpha }\oplus M_9\oplus {\left(1\right)}^2\oplus {\left(-1\right)}^2$	$A_{\alpha }^{\frac{v}{2}-2}{A}_{\alpha -1}{D}_{\frac{\alpha +2}{2}}{D}_{\frac{\alpha +2}{2}}{T}_{\frac{v}{2}-1}$
	if $\beta =v-2$	if $\beta =v-2$
(o)	$M_1^{\alpha }\oplus M_6\oplus \left({\omega }^{\frac{v}{2}-1}\right)\oplus \left({\omega }^{-\left(\frac{v}{2}-1\right)}\right)\oplus {\left(1\right)}^2$	$A_{\alpha }^{\frac{\beta }{2}}{A}_{\alpha -1}^{\frac{v}{2}-1-\frac{\beta }{2}}{D}_{\frac{\alpha }{2}}{D}_{\frac{\alpha +2}{2}}{T}_{\frac{v}{2}-1}$
	if $\beta \ne v-2$	if $\beta \ne v-2$
	$M_1^{\alpha }\oplus M_9\oplus {\left(1\right)}^2\oplus {\left(-1\right)}^2$	$A_{\alpha }^{\frac{v}{2}-2}{A}_{\alpha -1}{D}_{\frac{\alpha +2}{2}}{D}_{\frac{\alpha +2}{2}}{T}_{\frac{v}{2}-1}$
	if $\beta =v-2$	if $\beta =v-2$
(p)	$M_1^{\alpha }\oplus {\left(-1\right)}^2$	$A_{\alpha -1}^{\frac{v}{2}-1}{D}_{\frac{\alpha }{2}}{D}_{\frac{\alpha +2}{2}}{T}_{\frac{v}{2}-1}$

Table 8. Table 8: The possible exceptions to d u ​ ( G s . c . ) = d u ​ ( G a . ) subscript 𝑑 𝑢 subscript 𝐺 formulae-sequence 𝑠 𝑐 subscript 𝑑 𝑢 subscript 𝐺 𝑎 d_{u}(G_{s.c.})=d_{u}(G_{a.}) for G 𝐺 G of classical type

$G$	Cases	$d_u\left(G_{s.c}\right)-d_u\left(G_{a.}\right)$
$A_{\ell }$	$p\ne 2$, $u$ even, $h=zu$ and $z$ odd	$\le 2$
$C_{\ell }$	$p\ne 2$ and $u$ even	$\le 2\lceil \frac{z}{2}\rceil$
$B_{\ell }$	$p\ne 2$, $u\equiv 2mod4$, $h=zu$, and $z\equiv u/2mod4$ or $z\equiv 2mod4$	$\le 2$
	$p\ne 2$, $u\equiv 4mod8$, $h=zu$ and $z$ odd
$D_{\ell }$	$p\ne 2$, $u=2$, $h=zu$ and $z\equiv u/2mod4$	$\le 4$
	$p\ne 2$, $u\equiv 2mod4$, $u>2$, $h=zu$ and $z\equiv u/2mod4$	$\le 2$
	$p\ne 2$, $u\equiv 2mod4$, $e=u-2\ne 0$ and $z\equiv 1mod4$
	$p\ne 2$, $u\equiv 4mod8$, $z$ is odd, and $h=zu$

Equations

$$G_{[u]}=\{g\in G:g^u=1\}$$

$$d_u(G_{a.})\le d_u(G)\le d_u(G_{s.c.}).$$

$$d_u(G_{s.c.})=d_u(G_{a.})$$

$$d_u(G_{s.c.})-d_u(G_{a.})=2.$$

$$d_u(G_{s.c.})=d_u(G_{a.})$$

$$d_u(G_{s.c.})-d_u(G_{a.})=2.$$

$$d_u(G_{s.c.})=d_u(G_{a.})$$

$$d_u(G_{s.c.})-d_u(G_{a.})=2\lceil \frac{z}{2}\rceil .$$

$$d_u(G_{s.c.})=d_u(G_{a.})$$

$$d_u(G_{s.c.})-d_u(G_{a.})=4.$$

$$d_u(G_{s.c.})-d_u(G_{a.})=2.$$

$$d_2(G_{s.c.})-d_2(G_{a.})=6\text{and}{d}_u(G_{s.c.})-d_u(G_{a.})=2\text{for}u\in \{6,10,14,18\}.$$

$$d_{u}(G):\begin{array}[]{l}\mathbb{N}\rightarrow\mathbb{N}\\ u\mapsto d_{u}(G).\end{array}$$

$$T=T_{a,b,c}=⟨x,y,z:x^a=y^b=z^c=xyz=1⟩.$$

$$j_u(G)=\dim G_{[u]}$$

$$C_{C_G(y{)}^0}(x)=C_{H_1}(x_1)\dots C_{H_r}(x_r)T^{\prime }.$$

$$\dim C_G(g)=\dim C_G(g{)}^0\text{and}\dim C_{C_G(y{)}^0}(x)=\dim C_{C_G(y{)}^0}(x{)}^0.$$

$$\begin{array}[]{llll}d_{u}(G)&=&\min&d_{q}(C_{G}(y)^{0}).\\ &&y\in G_{[v]}&\\ \end{array}$$

$$d_u(G)=d_q(C_G(y{)}^0).$$

$$d_q(C_G(y{)}^0)=\mathrm{rank}(T^{\prime })+\sum _{i=1}^r{d}_q(H_i).$$

$$\begin{array}[]{llll}d_{u}(G)&=&\min&{\rm rank}(T^{\prime})+d_{q}(H_{1})+\dots+d_{q}(H_{r}).\\ &&C_{G}(y)^{0}=\prod_{i=1}^{r}H_{i}T^{\prime}&\\ &&y\in G_{[v]}&\\ \end{array}$$

$$\begin{array}[]{lllll}d_{u}(G)&=&\min&\min&\dim C_{C_{G}(y)^{0}}(x).\\ &&y\in G_{[v]}&x\in(C_{G}(y)^{0})_{[q]}&\end{array}$$

$$d_u(G)=\dim C_{C_G(y{)}^0}(x).$$

$$\dim C_{C_G(y{)}^0}(x)=\mathrm{rank}(T^{\prime })+\sum _{i=1}^r\dim C_{H_i}(x_i).$$

$$d_q(C_G(y{)}^0)$$

$$G_{g,k}=\{x\in G:g^{-1}{x}^{-1}gx=k\}.$$

Full text

1. 1 Introduction

2. 2 Preliminary results on centralizers

3. 3 Proofs of Propositions 2 and 3

4. 4 Proof of Theorem 4 for $G$ of exceptional type

5. 5 Some properties of ${\rm Spin}_{n}(K)$

6. 6 Upper bounds for $d_{u}(G)$ for $G$ of classical type

7. 6.1 $G$ is of type $A_{\ell}$

8. 6.2 $G$ is of type $C_{\ell}$

9. 6.3 $G$ is of type $B_{\ell}$

10. 6.4 $G$ is of type $D_{\ell}$

11. 7 Proofs of Theorems 4 and 5

12. 8 Proof of Proposition 7

13. 9 Proof of Theorem 9

14. 10 Proof of Proposition 10

15. 11 Some tables

1 Introduction

Let $G$ be a reductive algebraic group defined over an algebraically closed field $K$ of characteristic $p$ (possibly equal to 0), $C$ be a conjugacy class of $G$ and $u$ be a positive integer. In 2007 Guralnick [4] proved the following result:

Theorem 1**.**

[4, Theorem 1.1]**. Given a reductive algebraic group $G$, a conjugacy class $C$ of $G$ and a positive integer $u$, the set $\{g\in G:g^{u}\in C\}$ is a finite union of conjugacy classes of $G$.

In this paper, we concentrate our attention to the case where $G$ is connected and $C=\{1\}$ is the trivial conjugacy class of $G$. For a positive integer $u$, we let

[TABLE]

be the subvariety of $G$ consisting of elements of $G$ of order dividing $u$ and set $j_{u}(G)=\dim G_{[u]}$. We also let $d_{u}(G)$ be the minimal dimension of a centralizer in $G$ of an element of $G$ of order dividing $u$. We are merely interested in determining $j_{u}(G)$ for every positive positive integer $u$ (when $G$ is a simple algebraic group).

For completeness, we begin by proving Guralnick’s result in the case where $G$ is connected and $C=\{1\}$. In the statement below, given $g\in G$, we let $g^{G}$ denote the conjugacy class of $g$ in $G$.

Proposition 2**.**

Let $G$ be a connected reductive algebraic group defined over an algebraically closed field $K$ of characteristic $p$. Let $u$ be a positive integer. Then the number of conjugacy classes of $G$ of elements of order dividing $u$ is finite. In particular $G_{[u]}$ is a finite union of conjugacy classes of $G$. Moreover $\dim G_{[u]}=\max_{g\in G_{[u]}}\dim g^{G}$ and ${\rm codim}\ G_{[u]}=d_{u}(G)$.

Given a simple algebraic group $G$ defined over an algebraically closed field $K$ of characteristic $p$, we denote by $G_{s.c.}$ (respectively, $G_{a.}$) the simple algebraic group over $K$ of simply connected (respectively, adjoint) type having the same Lie type and Lie rank as $G$. We prove the following result partially proved in [7, Theorem 3.11]:

Proposition 3**.**

Let $G$ be simple algebraic group defined over an algebraically closed field $K$ of characteristic $p$. Let $u$ be a positive integer. Then

[TABLE]

In [7], given a positive integer $u$ and a simple algebraic group $G$ defined over an algebraically closed field $K$ of characteristic $p$, Lawther gives a lower bound for $d_{u}(G)$. Moreover, he proves that this lower bound is equal to $d_{u}(G)$ in the case where $G$ is of adjoint type. We establish the following result completing Lawther’s result. Recall that the Coxeter number of a simple algebraic group is defined to be the quotient of the number of roots in the root system associated to $G$ by the number of such roots which are simple. Concretely, $h=\ell+1$, $2\ell$, $2\ell$, $2\ell-2$, $6$, $12$, $12$, $18$ or $30$ according respectively as $G$ is of type $A_{\ell}$, $B_{\ell}$, $C_{\ell}$, $D_{\ell}$, $G_{2}$, $F_{4}$, $E_{6}$. $E_{7}$ or $E_{8}$.

Theorem 4**.**

Let $G$ be a simple algebraic group defined over an algebraically closed field $K$ of characteristic $p$. Let $h$ be the Coxeter number of $G$ and $u$ be a positive integer. Write $h=zu+e$ and $u=qv$, where $z$ and $e$ are nonnegative integers with $0\leq e\leq u-1$ and $q$, $v$ are coprime positive integers such that $q$ is a power of $p$ and $p$ does not divide $v$. The following assertions hold.

1. (i)

If $G$ is of type $A_{\ell}$ then

[TABLE]

except if $p\neq 2$, $u$ is even, $h=zu$ and $z$ is odd in which case

[TABLE] 2. (ii)

If $G$ is of type $B_{\ell}$ then

[TABLE]

except if

1. (a)

$p\neq 2$, $u\equiv 2\mod 4$, $h=zu$, and $z\equiv u/2\mod 4$ or $z\equiv 2\mod 4$ in which case

[TABLE] 2. (b)

$p\neq 2$, $u\equiv 4\mod 8$, $h=zu$ and $z$ is odd in which case

[TABLE] 3. (iii)

If $G$ is of type $C_{\ell}$ then

[TABLE]

except if $p\neq 2$ and $u$ is even in which case

[TABLE] 4. (iv)

If $G$ is of type $D_{\ell}$ then

[TABLE]

except if

1. (a)

$p\neq 2$, $u=2$, $h=zu$ and $z\equiv u/2\mod 4$ in which case

[TABLE] 2. (b)

$p\neq 2$, $u\equiv 2\mod 4$, $u>2$, $h=zu$ and $z\equiv u/2\mod 4$ in which case

[TABLE] 3. (c)

$p\neq 2$, $u\equiv 2\mod 4$, $e=u-2\neq 0$ and $z\equiv 1\mod 4$ in which case

[TABLE] 4. (d)

$p\neq 2$, $u\equiv 4\mod 8$, $z$ is odd and $h=zu$ in which case

[TABLE] 5. (v)

If $G$ is of exceptional type then $d_{u}(G_{s.c.})=d_{u}(G_{a.})$ except if $G$ is of type $E_{7}$, $p\neq 2$ and $u\in\{2,6,10,14,18\}$ in which case

[TABLE]

We note that for $G$ of exceptional type the only cases where the simply connected and the adjoint groups are not abstractly isomorphic is when $G$ is of type $E_{6}$ or $E_{7}$ and $p\neq 3$ or $p\neq 2$ respectively.

Recall that if a simple algebraic group $G$ over an algebraically closed field $K$ of characteristic $p$ is not of simply connected type nor of adjoint type then either $G$ is abstractly isomorphic to ${\rm SL}_{n}(K)/C$ where $C\leqslant Z({\rm SL}_{n}(K))$, or $G$ is of type $D_{\ell}$, $p\neq 2$ and $G$ is abstractly isomorphic to ${\rm SO}_{2\ell}(K)$ or a half-spin group ${\rm HSpin}_{2\ell}(K)$ where $\ell$ is even in the latter case. For simple algebraic groups that are neither of simply connected nor adjoint type we obtain the following result.

Theorem 5**.**

Let $G$ be a simple algebraic group, neither of simply connected nor adjoint type, defined over an algebraically closed field $K$ of characteristic $p$. Let $u$ be a positive integer. Write $u=qv$ where $q$ and $v$ are coprime positive integers such that $q$ is a power of $p$ and $p$ does not divide $v$. Then the following assertions hold.

1. (i)

If $d_{u}(G_{s.c.})=d_{u}(G_{a.})$ then $d_{u}(G)=d_{u}(G_{a.})$ 2. (ii)

Suppose $d_{u}(G_{s.c.})\neq d_{u}(G_{a.})$.

1. (a)

If $G={\rm SL}_{\ell+1}(K)/C$ where $C\leq Z({\rm SL}_{\ell+1}(K))$, then $d_{u}(G)=d_{u}(G_{a.})$ if $C$ has an element of order $2$, otherwise $d_{u}(G)=d_{u}(G_{{s.c.}})$. 2. (b)

If $G={\rm SO}_{2\ell}(K)$ where $p\neq 2$, then $d_{u}(G)=d_{u}(G_{a.})$. 3. (c)

If $G={\rm HSpin}_{2\ell}(K)$ where $p\neq 2$ and $\ell$ is even, then $d_{u}(G)=d_{u}(G_{s.c.})$.

In [7] Lawther showed that if $G$ is a simple algebraic group of rank $\ell$ with Coxeter number $h$ over an algebraically closed field and $u$ is a positive integer with $u\geq h$ then $d_{u}(G)=\ell$. As a corollary of Theorems 4 and 5 we obtain the following result:

Corollary 6**.**

Let $G$ be a simple algebraic group of rank $\ell$ over an algebraically closed field $K$ of characteristic $p$. Let $h$ be the Coxeter number of $G$ and let $u$ be a positive number. Then

1. (i)

If $u>h$ then $d_{u}(G)=\ell$. 2. (ii)

Suppose $u=h$ then $d_{u}(G)=\ell$ if $G$ is (abstractly isomorphic to a group) of adjoint type, or $p=2$, or $h$ is odd (and then $G$ is of type $A$), or $G={\rm SL}_{\ell+1}(K)/C$ and $C$ has an element of order $2$ where $C\leq Z({\rm SL}_{\ell+1}(K))$, or $G\cong{\rm SO}_{2\ell}(K)$ and $p\neq 2$, or $G$ is of type $B$ or $D$ and $h\equiv 0\mod 8$, or $G=E_{6}(K)$, otherwise $d_{u}(G)=\ell+2$.

Given a simple algebraic group $G$ over an algebraically closed field, we also derive a further property of

[TABLE]

Proposition 7**.**

Let $G$ be a simple algebraic group over an algebraically closed field $K$ of characteristic $p$. Then

[TABLE]

is a decreasing function of $u$.

We now apply our results to the study of finite quasisimple images of triangle groups. Recall that a finite group is quasisimple if it is perfect and the quotient by its centre is simple.

Every finite quasisimple group being 2-generated, given a finite quasisimple group $G_{0}$ and a triple $(a,b,c)$ of positive integers, a natural question to consider is whether $G_{0}$ can be generated by two elements of orders dividing respectively $a$ and $b$ and having product of order dividing $c$. A finite group generated by two such elements is called an $(a,b,c)$-group. Equivalently, an $(a,b,c)$-group is a finite quotient of the triangle group $T=T_{a,b,c}$ with presentation

[TABLE]

When investigating the finite (nonabelian) quasisimple quotients of $T$, we can assume that $1/a+1/b+1/c<1$ as otherwise $T$ is either soluble or $T\cong T_{2,3,5}\cong{\rm Alt}_{5}$ (see [2]). The group $T$ is then a hyperbolic triangle group. Without loss of generality, we will further assume that $a\leq b\leq c$ and call $(a,b,c)$ a hyperbolic triple of integers. (Indeed, $T_{a,b,c}\cong T_{a^{\prime},b^{\prime},c^{\prime}}$ for any permutation $(a^{\prime},b^{\prime},c^{\prime})$ of $(a,b,c)$.)

Recall that a finite quasisimple group $G_{0}$ of Lie type occurs as the derived subgroup of the fixed point group of a simple algebraic group $G$, when ${\rm char}(K)=p$ is prime, under a Steinberg endomorphism $F$, i.e. $G_{0}=(G^{F})^{\prime}$. We use the standard notation $G_{0}=(G^{F})^{\prime}=G(q)$ where $q=p^{r}$ for some positive integer $r$. (We include the possibility that $G(q)$ is of twisted type.)

Given a simple algebraic group $G$ and a triple $(a,b,c)$ of positive integers, we say that $(a,b,c)$ is rigid for $G$ if the sum $j_{a}(G)+j_{b}(G)+j_{c}(G)$ is equal to $2\dim G$. When the latter sum is less (respectively, greater) than $2\dim G$, we say that $(a,b,c)$ is reducible (respectively, nonrigid) for $G$.

In [10] we conjectured the following finiteness result in the rigid case (first formulated for hyperbolic triples $(a,b,c)$ of primes):

Conjecture 8**.**

Let $G$ be a simple algebraic group defined over an algebraically closed field $K$ of prime characteristic $p$ and let $(a,b,c)$ be a rigid triple of integers for $G$. Then there are only finitely many quasisimple groups $G(p^{r})$ (of the form $(G^{F})^{\prime}$ where $F$ is a Steinberg endomorphism) that are $(a,b,c)$-generated.

In [6, Theorem 1.7], Larsen, Lubotzky and the author proved that Conjecture 8 holds except possibly if $p$ divides $abcd$ where $d$ is the determinant of the Cartan matrix of $G$. In a future paper [5] we will make further progress on Conjecture 8 by, in particular, completely settling it in the case where the finite group $(G^{F})^{\prime}$ is simple. A crucial ingredient therein is the classification of the hyperbolic triples $(a,b,c)$ of integers as reducible, rigid or nonrigid for a given simple algebraic group $G$. We determine this classification below.

Theorem 9**.**

The following assertions hold.

1. (i)

The reducible hyperbolic triples $(a,b,c)$ of integers for simple algebraic groups $G$ of simply connected or adjoint type are exactly those given in Table 1. In particular there are no reducible hyperbolic triples of integers for simple algebraic groups of adjoint type. 2. (ii)

The rigid hyperbolic triples of integers $(a,b,c)$ for simple algebraic groups $G$ of simply connected type are exactly those given in Table 2. 3. (iii)

The rigid hyperbolic triples of integers $(a,b,c)$ for simple algebraic groups $G$ of adjoint type are exactly those given in Table 3. 4. (iv)

The classification of the reducible and the rigid hyperbolic triples of integers for ${\rm SO}_{n}(K)$ is the same as for ${\rm PSO}_{n}(K)$. 5. (v)

The classification of the reducible and the rigid hyperbolic triples of integers for ${\rm HSpin}_{2\ell}(K)$ is the same as for ${\rm Spin}_{2\ell}(K)$. 6. (vi)

If $C\leqslant Z({\rm SL}_{n}(K))$ contains an involution then the classification of the reducible and the rigid hyperbolic triples of integers for ${\rm SL}_{n}(K)/C$ is the same as for ${\rm PSL}_{n}(K)$. Otherwise, the classification of the reducible and the rigid hyperbolic triples of integers for ${\rm SL}_{n}(K)/C$ is the same as for ${\rm SL}_{n}(K)$.

As Theorem 9 gives the list of the rigid triples of integers for a given simple algebraic group, it puts Conjecture 8 into a very concrete context. Concerning the reducible case, we prove the following nonexistence result.

Proposition 10**.**

Let $G$ be a simple algebraic group defined over an algebraically closed field $K$ of prime characteristic $p$ and let $(a,b,c)$ be a reducible triple of integers for $G$. Then a quasisimple group $G(p^{r})$ (of the form $(G^{F})^{\prime}$ where $F$ is a Steinberg endomorphism) is never an $(a,b,c)$-group.

Theorem 9 which gives inclusively the list of the reducible triple of integers for a given simple algebraic group is one of the ingredients of the proof of Proposition 10. As an immediate corollary of Theorem 9 and Proposition 10 we obtain the following result.

Corollary 11**.**

If $(G,p,(a,b,c))$ is as in Table 1 then $(G^{F})^{\prime}=G(p^{r})$ is never an $(a,b,c)$-group.

Unless otherwise stated, we let $G$ be a connected reductive algebraic group defined over an algebraically closed field $K$ of characteristic $p$ with maximal torus $T$. If $T^{\prime}$ is a torus of $G$ of dimension $r$ we sometimes write $T^{\prime}=T_{r}$.

Given a positive integer $u$, we let $G_{[u]}$ be the subvariety of $G$ consisting of elements of order dividing $u$, we set

[TABLE]

and we let $d_{u}(G)$ be the minimal dimension of a centralizer in $G$ of an element of $G$ of order dividing $u$.

For $G$ simple, we fix some more notation as follows. We let $\Phi$ be the root system of $G$ with respect to $T$ and set $\Pi=\{\alpha_{1},\dots,\alpha_{\ell}\}\subset\Phi$ to be a set of simple roots of $G$, where $\ell$ is the rank of $G$. We let $h=|\Phi|/{\rm rank}(G)=|\Phi|/\ell$ be the Coxeter number of $G$. Recall that $h=\ell+1$, $2\ell$, $2\ell$, $2\ell-2$, $6$, $12$, $12$, $18$ or $30$ according respectively as $G$ is of type $A_{\ell}$, $B_{\ell}$, $C_{\ell}$, $D_{\ell}$, $G_{2}$, $F_{4}$, $E_{6}$. $E_{7}$ or $E_{8}$. Also we denote by $G_{s.c.}$ (respectively, $G_{a.}$) the simple algebraic group over $K$ of simply connected (respectively, adjoint) type having the same Lie type and Lie rank as $G$.

Unless otherwise stated given a positive integer $u$, we also write:

$u=qv$ where $q$ and $v$ are coprime positive integers such that $q$ is a power of $p$ and $p$ does not divide $v$,

$h=zu+e$ where $z,e$ are nonnegative integers such that $0\leq e\leq q-1$,

$h=\alpha v+\beta$ where $\alpha,\beta$ are nonnegative integers such that $0\leq\alpha\leq v-1$,

$\alpha=\gamma q+\delta$ where $\gamma,\delta$ are nonnegative integers such that $0\leq\delta\leq q-1$.

An easy check yields $\gamma=z$. Note that if $e=0$ then $\beta=\delta=0$. Furthermore, if $\beta=0$ then $\delta=0$ if and only if $e=0$.

Finally for an nonnegative integer $r$, we let $\epsilon_{r}\in\{0,1\}$ be [math] if $r$ is even, otherwise $\epsilon_{r}=1$, and set $\sigma_{r}\in\{0,1\}$ to be 1 if $r=0$, otherwise $\sigma_{r}=0$.

The outline of the paper is as follows. In §2 we give some preliminary results on centralizers. In §3 we give detailed proofs for Propositions 2 and 3. In §4, given a positive integer $u$, we determine $d_{u}(G)$ for $G$ a simple simply connected algebraic group of exceptional type over an algebraically closed field $K$. This establishes Theorem 4 for $G$ of exceptional type. In §5 we recall some properties of the spin groups and determine when a preimage under the canonical map ${\rm Spin}_{n}(K)\rightarrow{\rm SO}_{n}(K)$ of a semisimple element of ${\rm SO}_{n}(K)$ of a given order is also of that order. In §6 given a positive integer $u$, we determine precise upper bounds for $d_{u}(G)$ where $G$ is a simple algebraic group of classical type over an algebraically closed field. In §7 given a positive integer $u$, we determine $d_{u}(G)$ for $G$ of classical type, completing the proofs of Theorems 4 and 5. In §8 we prove that for a given simple algebraic group $G$ over an algebraically closed field, $d_{u}(G)$, seen as a function of $u$, is decreasing. This establishes Proposition 7. In §9 we classify the reducible and the rigid hyperbolic triples of integers for simple algebraic groups, establishing Theorem 9. In §10 we prove Proposition 10. Finally in §11 we collect Tables 4-8 which appear later in the paper.

Acknowledgements. The author thanks the MARIE CURIE and PISCOPIA research fellowship scheme and the University of Padova for their support. The research leading to these results has received funding from the European Comission, Seventh Framework Programme (FP7/2007-2013) under Grant Agreement 600376.

2 Preliminary results on centralizers

In this section, unless otherwise stated, $G$ denotes a connected reductive group defined over an algebraically closed field $K$ of characteristic $p$. Given a positive integer $u$, recall that $d_{u}(G)$ is defined to be the minimal dimension of a centralizer in $G$ of an element of $G$ of order dividing $u$. We first recall some generalities on the centralizer in $G$ of an element of $G$.

Lemma 2.1**.**

Let $G$ be a connected reductive group defined over an algebraically closed field $K$ of characteristic $p$. Let $g$ be an element of $G$ with Jordan decomposition $g=xy$ where $x$ is unipotent and $y$ is semisimple. Let $H=C_{G}(y)^{0}$. The following assertions hold.

1. (i)

We have $x\in C_{G}(y)^{0}$, $C_{G}(g)=C_{C_{G(y)}}(x)$ and $C_{G}(g)^{0}=C_{C_{G}(y)^{0}}(x)^{0}$. 2. (ii)

The group $H=C_{G}(y)^{0}$ is reductive and so $H=H_{1}\cdots H_{r}T^{\prime}$ for some nonnegative integer $r$ and simple groups $H_{i}$ and central torus $T^{\prime}$. Morever each $H_{i}$ is closed and normal in $H$, $[H_{i},H_{j}]=1$ for $i\neq j$, and $H_{i}\cap H_{1}\dots H_{i-1}H_{i+1}\cdots H_{r}$ is finite for each $i$. 3. (iii)

Write $x=x_{1}\cdots x_{r}t$ where $x_{i}\in H_{i}$ and $t\in T^{\prime}$. Then

[TABLE] 4. (iv)

$\dim C_{G}(g)=\dim C_{C_{G}(y)^{0}}(x)={\rm rank}(T^{\prime})+\sum_{i=1}^{r}\dim C_{H_{i}}(x_{i}).$**

Proof.

We first consider part (i). Since $G$ is connected reductive we have $x\in C_{G}(y)^{0}$ - see for example [9, Proposition 14.7]. It is well-known that $C_{G}(g)=C_{C_{G(y)}}(x)$. As $C_{G}(g)\leqslant C_{G}(y)$, we have $C_{G}(g)^{0}\leqslant C_{G}(y)^{0}$. It now follows that $C_{G}(g)^{0}\leqslant C_{C_{G}(y)^{0}}(x)$ and so $C_{G}(g)^{0}=C_{C_{G}(y)^{0}}(x)^{0}$.

We consider part (ii). Since $G$ is connected reductive and $y$ is a semisimple element of $G$, $C_{G}(y)^{0}$ is connected reductive - see for example [9, Theorem 14.2]. Since $H=C_{G}(y)^{0}$ is connected reductive, we have $H=[H,H]Z(H)^{0}$ where $Z(H)^{0}$ is a central torus and $[H,H]$ is semisimple - see for example [9, Corollary 8.22]. Part (ii) follows - see for example [9, Theorem 8.21]. Part (iii) is now an easy consequence of part (ii).

We now consider part (iv). We have

[TABLE]

Part (iv) now follows from parts (i)-(iii).

∎

We now give some properties of $d_{u}(G)$. In the statement below, we adopt the notation of Lemma 2.1(ii) for the connected component of the centralizer of a semisimple element in a connected reductive algebraic group.

Proposition 2.2**.**

Let $G$ be a connected reductive algebraic group over an algebraically closed field $K$ of characteristic $p$. Let $u$ be a positive integer. Write $u=qv$ where $u$ and $v$ are coprime positive integers with $q$ a power of $p$ and $p$ does not divide $v$. Then the following assertions hold.

1. (i)

[TABLE] 2. (ii)

Let $g=xy$ be the Jordan decomposition of an element $g\in G$ where $x$ and $y$ are respectively the unipotent and semisimple parts of $g$. If $g$ has order dividing $u$ and $d_{u}(G)=\dim C_{G}(g)$, then

[TABLE] 3. (iii)

If $y$ is a semisimple element of $G$ with $C_{G}(y)^{0}=H_{1}\dots H_{r}T^{\prime}$ (as in Lemma 2.1(ii)) then

[TABLE] 4. (iv)

[TABLE]

Proof.

We first consider part (i). By definition $d_{u}(G)=\min_{g\in G_{[u]}}\ \dim C_{G}(g)$. Let $g$ be any element of $G_{[u]}$. Write $g=xy$ where $x$ and $y$ are respectively the unipotent and semisimple parts of $g$. Clearly $x\in G_{[q]}$ and $y\in G_{[v]}$. Following Lemma 2.1(i), we in fact have $x\in(C_{G}(y)^{0})_{[q]}$ and $C_{G}(g)^{0}=C_{C_{G}(y)^{0}}(x)^{0}$. It follows that

[TABLE]

This establishes part (i).

We now consider part (ii). Note that $x\in(C_{G}(y)^{0})_{[q]}$ and $y\in G_{[v]}$. Since by assumption $d_{u}(G)=\dim C_{G}(g)$ and by Lemma 2.1(iv) $\dim C_{G}(g)=\dim C_{C_{G}(y)^{0}}(x)$, we obtain

[TABLE]

By part (i) $d_{u}(G)=\min_{l\in G_{[v]}}d_{q}(C_{G}(l)^{0})$ and so $d_{q}(C_{G}(y)^{0})\geq d_{u}(G)$. However if $d_{q}(C_{G}(y)^{0})>d_{u}(G)$ then $\dim C_{C_{G}(y)^{0}}(x)>d_{u}(G)$, contradicting (2.1). Hence $d_{u}(G)=d_{q}(C_{G}(y)^{0})$ as claimed.

We now consider part (iii). Recall that by Lemma 2.1(ii), $C_{G}(y)^{0}$ is connected reductive. For $x\in(C_{G}(y)^{0})_{[q]}$, write $x=x_{1}\dots x_{r}t$ where $x_{i}\in H_{i}$ for $1\leq i\leq r$ and $t\in T^{\prime}$. By Lemma 2.1(iv) we have

[TABLE]

Hence

[TABLE]

where to obtain the final equality we used Lemma 2.1(ii)-(iii). Part (iii) follows.

Finally part (iv) follows from parts (i) and (iii). ∎

The result below is the main ingredient in the proof of Proposition 3.

Lemma 2.3**.**

Let $G$ be any group, $C$ a normal subgroup of $G$, and $\phi:G\rightarrow G/C$ be the canonical surjective map. Let $g,k$ be any elements of $G$ and set

[TABLE]

The following assertions hold.

1. (i)

Either $G_{g,k}=\emptyset$ or $G_{g,k}$ is a coset of $C_{G}(g)$. 2. (ii)

We have $C_{G/C}(gC)=\bigcup_{k\in C}\phi(G_{g,k})$. 3. (iii)

Suppose $G$ is a simple algebraic group defined over an algebraically closed field $K$ of characteristic $p$ and $\phi$ is an isogeny. Then

[TABLE]

In particular, $\dim C_{G/C}(gC)=\dim C_{G}(g)$. Moreover if $q$ is a power of $p$ and $g$ is a semisimple element of $G$ then

[TABLE]

Proof.

We first consider part (i). Note that if $g$ and $gk$ are not conjugate then $G_{g,k}=\emptyset$. Suppose that $g$ and $gk$ are conjugate, say $l^{-1}gl=gk$ for some $l\in G$.

Let $x$ be any element of $G_{g,k}$. Then

[TABLE]

Hence $lx^{-1}g(lx^{-1})^{-1}=g$ and $xl^{-1}\in C_{G}(g)$. It follows that $x\in C_{G}(g)l$ and so $G_{g,k}\subseteq C_{G}(g)l$.

Suppose now that $x\in C_{G}(g)l$. Then $x=yl$ for some $y\in C_{G}(g)$. Hence

[TABLE]

and so $x\in G_{g,k}$. It follows that $C_{G}(g)l\subseteq G_{g,k}$. We obtain $G_{g,k}=C_{G}(g)l$, proving part (i).

We now consider part (ii). Suppose first that $xC$ is any element of $\phi(G_{g,k})$ where $k\in C$. Without loss of generality, $x\in G_{g,k}$. Hence $x^{-1}gx=gk$. Since $k\in C$, we obtain $(xC)^{-1}\cdot gC\cdot xC=gC$ and so $xC\in C_{G/C}(gC)$. Hence $\bigcup_{k\in C}\phi(G_{g,k})\subseteq C_{G/C}(gC)$.

Suppose now that $xC\in C_{G/C}(gC)$. Then $(xC)^{-1}\cdot gC\cdot xC=gC$ and so $(gC)^{-1}\cdot(xC)^{-1}\cdot gC\cdot xC=C$. It follows that there exists $k\in C$ such that $g^{-1}x^{-1}gx=k$ and so $x\in G_{g,k}$. Hence $xC\in\phi(G_{g,k})$ with $k\in C$. This shows that $C_{G/C}(gC)\subseteq\bigcup_{k\in C}\phi(G_{g,k})$. We obtain $C_{G/C}(gC)=\bigcup_{k\in C}\phi(G_{g,k})$, proving part (ii).

Finally we consider part (iii). As $G$ is simple and $\phi$ is an isogeny, ${\rm ker}\ \phi=C$ is a finite central subgroup of $G$. Also following part (ii), $\phi(C_{G}(g))=C_{G}(g)/C$ is a subgroup of $C_{G/C}(gC)$ of finite index and so $C_{G/C}(gC)^{0}\leqslant(C_{G}(g)/C)^{0}$. On the other hand, as $C_{G}(g)/C\leqslant C_{G/C}(gC)$, we have $(C_{G}(g)/C)^{0}\leqslant C_{G/C}(gC)^{0}$. Hence

[TABLE]

Note that as $C_{G}(g)^{0}$ is connected so is the factor group $C_{G}(g)^{0}/(C\cap C_{G}(g)^{0})$ - see for example [9, Proposition 1.10]. By the second isomorphism theorem, $C_{G}(g)^{0}/(C\cap C_{G}(g)^{0})$ can be seen as a connected subgroup of $C_{G}(g)/C$. As this subgroup is of finite index, we obtain

[TABLE]

As $C$ is a finite group, it now follows from (2.2) that $\dim C_{G/C}(gC)=\dim C_{G}(g)$. To conclude suppose that $g$ is a semisimple element of $G$. By Lemma 2.1 $C_{G}(g)^{0}$ is a connected reductive group. By [7] given a simple algebraic group $H$ defined over $K$, $d_{q}(H)$ is independent of the isogeny type of $H$ (see Lemma 2.4 below). It follows that $d_{q}(C_{G}(g)^{0}/(C\cap C_{G}(g)^{0}))=d_{q}(C_{G}(g)^{0})$ and so by (2.2) $d_{q}(C_{G}(g)^{0})=d_{q}(C_{G/C}(gC)^{0}),$ as required. ∎

For matter of clarity we record [7, Lemma 2.5 and Corollary 2.6]:

Lemma 2.4**.**

Let $G$ be a simple algebraic group defined over an algebraic closed field of characteristic $p$. Let $q$ be a power of $p$. Then $d_{q}(G)$ does not depend on the isogeny type of $G$. Furthermore, let $\alpha$ be a nonnegative integer and write $\alpha=\gamma q+\delta$ where $0\leq\delta\leq q-1$ and let $\varsigma\in\{0,1\}$. Then the following assertions hold:

1. (i)

$d_{q}(A_{\alpha-\varsigma})=\gamma^{2}q+(2\gamma+1)(\alpha-\varsigma-\gamma q+1)-1$. 2. (ii)

$d_{q}(B_{\lceil\frac{\alpha}{2}\rceil}-\varsigma\epsilon_{\alpha})=\frac{\gamma^{2}q}{2}+(2\gamma+1)(\lceil\frac{\alpha}{2}\rceil-\varsigma\epsilon_{\alpha}-\frac{\gamma q}{2})+\lceil\frac{\gamma}{2}\rceil\epsilon_{q}$. 3. (iii)

$d_{q}(C_{\lceil\frac{\alpha}{2}\rceil}-\varsigma\epsilon_{\alpha})=\frac{\gamma^{2}q}{2}+(2\gamma+1)(\lceil\frac{\alpha}{2}\rceil-\varsigma\epsilon_{\alpha}-\frac{\gamma q}{2})+\lceil\frac{\gamma}{2}\rceil\epsilon_{q}$. 4. (iv)

$d_{q}(D_{\lceil\frac{\alpha+1}{2}\rceil-\varsigma\epsilon_{\alpha+1})}=\frac{\gamma^{2}q}{2}+(2\gamma+1)(\lceil\frac{\alpha+1}{2}\rceil-\varsigma\epsilon_{\alpha+1}-\frac{\gamma q}{2})+\lceil\frac{\gamma}{2}\rceil\epsilon_{q}-\gamma-\epsilon_{\gamma}+2\varsigma\epsilon_{\gamma(\alpha+1)}\sigma_{\alpha-\gamma q}$.

Corollary 2.5**.**

Let $m$ be a natural number. The following assertions hold:

1. (i)

If $m+1=\gamma q+\delta$ with $0\leq\gamma\leq q-1$, then $d_{q}(A_{m+1})-d_{q}(A_{m})=2\gamma+1$. 2. (ii)

If $2m+1=\gamma q+\delta$ with $0\leq\gamma\leq q-1$, then $d_{q}(B_{m+1})-d_{q}(B_{m})=2\gamma+1$. 3. (iii)

If $2m+1=\gamma q+\delta$ with $0\leq\gamma\leq q-1$, then $d_{q}(C_{m+1})-d_{q}(C_{m})=2\gamma+1$. 4. (iv)

If $2m=\gamma q+\delta$ with $0\leq\gamma\leq q-1$, then $d_{q}(D_{m+1})-d_{q}(D_{m})=2\gamma+1-2\epsilon_{\gamma}\sigma_{\delta}$.

Finally we record the following result:

Lemma 2.6**.**

[3, 6.17]**. Let $G$ be a simple algebraic group of rank $\ell$ over an algebraically closed field $K$ and let $x\in G$. Then

[TABLE]

3 Proofs of Propositions 2 and 3

In this section we prove Propositions 2 and 3. We first give a detailed proof of Proposition 2.

Proof of Proposition 2. Write $u=qv$ where $q$ and $v$ are coprime positive integers such that $q$ is a power of $p$ and $p$ does not divide $v$. Let $g$ be an element of $G$ of order dividing $u$. Let $g=xy$ be the Jordan decomposition of $g$ where $x$ and $y$ denote respectively the unipotent and semisimple parts of $g$. In particular the order of $x$ divides $q$ and that of $y$ divides $v$. As $G$ is connected, every semisimple element of $G$ lies in a maximal torus of $G$, see for example [9, Corollary 6.11]. Since all maximal tori of $G$ are conjugate, see for example [9, Corollary 6.5], it follows that there are only finitely many conjugacy classes of $G$ of semisimple elements of order dividing $v$. Let $y_{1},\dots,y_{r}$ be representatives of the conjugacy classes of $G$ of semisimple elements of order dividing $v$. For $1\leq i\leq r$, $C_{G}(y_{i})^{0}$ is a connected reductive algebraic group (see Lemma 2.1(ii)) and so by [8] $C_{G}(y_{i})^{0}$ has only finitely many unipotent classes. Let $x_{i,1},\dots,x_{i,t_{i}}$ be representatives of the unipotent classes of $C_{G}(y_{i})^{0}$. Now there exists $1\leq i\leq r$ such that $y$ is conjugate in $G$ to $y_{i}$. Let $k$ be an element of $G$ such that $kyk^{-1}=y_{i}$. Then

[TABLE]

Note that $kxk^{-1}\in C_{G}(y_{i})$. Indeed

[TABLE]

Since $G$ is a connected reductive group and $kxk^{-1}\cdot y_{i}$ is the Jordan decomposition of $kxk^{-1}\cdot y_{i}$, it follows that $kxk^{-1}$ is a unipotent element of $C_{G}(y_{i})^{0}$ - see Lemma 2.1(i). Hence there exist $l\in C_{G}(y_{i})^{0}$ and $1\leq j\leq t_{i}$ such that $l\cdot kxk^{-1}\cdot l^{-1}=x_{i,j}.$ Also as $l\in C_{G}(y_{i})^{0}$, we have $ly_{i}l^{-1}=y_{i}$. Hence

[TABLE]

and so $g$ is conjugate to $x_{i,j}y_{i}$. It follows that there are at most $\sum_{i=1}^{r}t_{i}$ conjugacy classes of elements of $G$ of order dividing $u$. It follows that $G_{[u]}$ consists of the finite union of conjugacy classes of $G$ of elements of order dividing $u$. Hence

[TABLE]

Since for any element $g\in G$, ${\rm codim}\ g^{G}=\dim C_{G}(g)$ - see for example [3, Proposition 1.5], we obtain

[TABLE]

We can now prove Proposition 3.

Proof of Proposition 3. First, by [7], we have $d_{u}(G_{a.})\leq d_{u}(G)$. It remains to show that $d_{u}(G)\leq d_{u}(G_{s.c.})$. Note that the result is trivial unless $G$ is neither of adjoint nor simply connected type. Let $\phi:G_{s.c.}\rightarrow G$ be an isogeny. Let $g$ be any element of $G_{s.c.}$ of order dividing $u$ such that $d_{u}(G_{s.c.})=\dim C_{G_{s.c.}}(g)$. Then $\phi(g)$ is an element of $G$ of order dividing $u$ and by Lemma 2.3 $\dim C_{G}(\phi(g))=\dim C_{G_{s.c.}}(g)$. Hence $\dim C_{G}(\phi(g))=d_{u}(G_{s.c.})$. Since $d_{u}(G)\leq\dim C_{G}(\phi(g))$ we obtain $d_{u}(G)\leq d_{u}(G_{s.c.})$. $\square$

4 Proof of Theorem 4 for $G$ of exceptional type

In this section, we determine $d_{u}(G)$ for $G$ a simple algebraic group of exceptional type defined over an algebraically closed field $K$ of characteristic $p$. In particular we prove Theorem 4 for $G$ of exceptional type (see Propositions 4.2 and 4.3 below). Recall that a simple algebraic group of exceptional type is either simply connected or adjoint. Following the result of Lawther [7] giving $d_{u}(G)$ for $G$ of adjoint type, we suppose that $G$ is of simply connected type. Without loss of generality, we assume that $G$ is of type $E_{6}$ or $E_{7}$ and $p\neq 3$ or $2$ respectively, as these are the only cases where the simply connected and the adjoint groups are not abstractly isomorphic.

Let $T$ be a maximal torus of $G$ with corresponding root system $\Phi$ and set $\Pi=\{\alpha_{1},\dots,\alpha_{\ell}\}\subset\Phi$ to be a set of simple roots of $G$, where $\ell$ is the rank of $G$. Note that $\ell=6$ or $\ell=7$ according respectively as $G=E_{6}$ or $G=E_{7}$.

Given a positive integer $u$, we write $u=qv$ where $q$ and $v$ are positive coprime integers with $q$ a power of $p$ and $p$ does not divide $v$.

Let $y$ be a semisimple element of $G$ of order dividing $v$, i.e. $y\in G_{[v]}$. Since every semisimple element of $G$ belongs to a maximal torus of $G$ and all maximal tori are conjugate, we can assume without loss of generality that $y\in T$. Now

[TABLE]

where $\Psi=\{\alpha\in\Phi:\alpha(y)=1\}$ and $U_{\alpha}$ is the root subgroup of $G$ corresponding to $\alpha$. Also we can write

[TABLE]

where for $l\in K^{*}=K\setminus\{0\}$ and $1\leq i\leq\ell$, $h_{\alpha_{i}}(l)$ is the image of $\begin{pmatrix}l&0\\ 0&l^{-1}\end{pmatrix}\in{\rm SL}_{2}(K)$ under the canonical map ${\rm SL}_{2}(K)\rightarrow\langle U_{\alpha_{i}},U_{-\alpha_{i}}\rangle$, $k$ is an element of $K^{*}$ of order $v$, and $0\leq c_{i}\leq v-1$ is an integer for all $1\leq i\leq\ell$.

Given $\alpha\in\Phi$ we then have

[TABLE]

where

[TABLE]

and $(,)$ denotes the inner product of the Euclidean space spanned by $\Phi$.

Lemma 4.1**.**

Let $G$ be a simple simply connected algebraic group of type $X=E_{6}$ or $E_{7}$ defined over an algebraically closed field $K$ of characteristic $p$. Let $u$ be a positive integer. Then $d_{u}(G)=d_{u}(G_{a.})$ except possibly if $G=E_{6}$, $p\neq 3$ and $u\equiv 0\mod 3$, or $G=E_{7}$, $p\neq 2$ and $u\equiv 0\mod 2$.

Proof.

Set $C=Z(G)$ and note that

[TABLE]

Also note that if $(X,p)\in\{(E_{6},3),(E_{7},2)\}$ then $G$ is abstractly isomorphic to $G_{a.}$ and so $d_{u}(G)=d_{u}(G_{a.})$ for every positive integer $u$. We therefore assume that $(X,p)\not\in\{(E_{6},3),(E_{7},2)\}$. Then $C=Z(G)=\langle c\rangle$ is cyclic group of order $3$ or $2$ according respectively as $X=E_{6}$ or $E_{7}$. Let $\phi:G\rightarrow G_{a.}$ be the surjective canonical map. Let $gC$ be an element of $G_{a.}$ of order dividing $u$ such that $\dim C_{G_{a.}}(gC)=d_{u}(G_{a.})$.

Suppose $u\neq 0\mod|C|$. Then $g\in G$ has order dividing $u$, and by Lemma 2.3 $\dim C_{G}(g)=\dim C_{G_{a.}}(gC)$. Hence $\dim C_{G}(g)=d_{u}(G_{a.})$ and $d_{u}(G)\leq d_{u}(G_{a.})$. It now follows from Proposition 3 that $d_{u}(G)=d_{u}(G_{a.})$. ∎

Proposition 4.2**.**

Suppose $G$ is a simple simply connected algebraic group of type $E_{6}$ defined over an algebraic closed field $K$ of characteristic $p$. Let $u$ be a positive integer. Then $d_{u}(G)=d_{u}(G_{a.})$.

Proof.

Note that by [7] $d_{u}(G)\geq d_{u}(G_{a.})$ and $d_{u}(G_{a})=6$ for $u\geq h=12$.

By Lemma 4.1 we can assume without loss of generality that $p\neq 3$ and $u\equiv 0\mod 3$. Write $u=qv$ where $q$ and $v$ are coprime positive integers such that $q$ is a power of $p$ and $p$ does not divide $v$.

We first suppose that $u\leq h=12$.

Assume $u=3$. Then $q=1$ and $u=v=3$. Let

[TABLE]

where $k\in K^{*}$ is an element of order 3. Then $y_{3}$ is a semisimple element of $G$ of order 3. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y_{3})^{0})=d_{1}(A_{2}A_{2}A_{2})=24$. Now by [7] $d_{u}(G_{a.})=24$. We now deduce from Proposition 2.2(i) that $d_{u}(G)=d_{u}(G_{a.})=24.$

Assume $u=6$. Then $q=1$ and $u=v=6$, or $q=p=2$ and $v=3$. Suppose first that $q=1$. Let

[TABLE]

where $k\in K^{*}$ is an element of order 6. Then $y_{6}$ is a semisimple element of $G$ of order 6. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y_{6})^{0})=d_{1}(A_{1}A_{1}A_{1}T_{3})=12$.

Suppose now that $q=2$ and $v=3$. Consider the semisimple element $y_{3}$ of $G$ of order 3 defined above. We have $C_{G}(y_{3})^{0}=A_{2}A_{2}A_{2}$ and $d_{q}(C_{G}(y_{3})^{0})=d_{2}(A_{2}A_{2}A_{2})=12$.

Now by [7] $d_{u}(G_{a.})=12$. We now deduce from Proposition 2.2(i) that $d_{u}(G)=d_{u}(G_{a.})=12.$

Assume $u=9$. Then $q=1$ and $u=v=9$. Let

[TABLE]

where $k\in K^{*}$ is an element of order 9. Then $y_{9}$ is a semisimple element of $G$ of order 9. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y_{9})^{0})=d_{1}(A_{1}T_{5})=8$. Now by [7] $d_{u}(G_{a.})=8$. We now deduce from Proposition 2.2(i) that $d_{u}(G)=d_{u}(G_{a.})=8.$

Assume $u=12$. Then $q=1$ and $u=v=12$, or $q=4$, $p=2$ and $v=3$. Suppose first that $q=1$. Let

[TABLE]

where $k\in K^{*}$ is an element of order 12. Then $y_{12}$ is a semisimple element of $G$ of order 12. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y_{12})^{0})=d_{1}(T_{6})=6$.

Suppose now that $q=4$, $p=2$ and $v=3$. Consider the semisimple element $y_{3}$ of $G$ of order 3 defined above. We have $C_{G}(y_{3})^{0}=A_{2}A_{2}A_{2}$ and $d_{q}(C_{G}(y_{3})^{0})=d_{4}(A_{2}A_{2}A_{2})=6$. By Proposition 2.2(i) we obtain $d_{12}(G)=d_{12}(G_{a.})=6$.

We now suppose that $u>h=12$. If $v\geq h$ then let

[TABLE]

where $k\in K^{*}$ is an element of order $v$. Then $y$ is a semisimple element of $G$ of order $v$. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y)^{0})=d_{q}(T_{6})=6$ and so by Proposition 2.2(i) $d_{u}(G)=6$.

We finally suppose that $u>h=12$ and $v<h$. As we are assuming that $u\equiv 0\mod 3$ and $p\neq 3$ we have $v\equiv 0\mod 3$. Hence $v\in\{3,6,9\}$.

Assume $v=3$. As $u>12$ we have $q\geq 5$. Now $y_{3}$ is a semisimple element of $G$ of order $3$ with $C_{G}(y_{3})^{0}=A_{2}A_{2}A_{2}$. As $q\geq 3$, we have $d_{q}(C_{G}(y_{3})^{0})=d_{q}(A_{2}A_{2}A_{2})=6$.

Assume $v=6$. As $u>12$ we have $q\geq 3$. Now $y_{6}$ is a semisimple element of $G$ of order $6$ with $C_{G}(y_{3})^{0}=A_{1}A_{1}A_{1}T_{3}$. As $q\geq 2$, we have $d_{q}(C_{G}(y_{6})^{0})=d_{q}(A_{1}A_{1}A_{1}T_{3})=6$.

Assume $v=9$. As $u>12$ we have $q\geq 2$. Now $y_{9}$ is a semisimple element of $G$ of order $9$ with $C_{G}(y_{9})^{0}=A_{1}T_{5}$. As $q\geq 2$, we have $d_{q}(C_{G}(y_{9})^{0})=d_{q}(A_{1}T_{5})=6$.

Hence in the case where $u>h$ and $v<h$, by Proposition 2.2(i), we get $d_{u}(G)=6.$ ∎

Proposition 4.3**.**

Suppose $G$ is a simple simply connected algebraic group of type $E_{7}$ defined over an algebraic closed field $K$ of characteristic $p$. Let $u$ be a positive integer. The following assertions hold.

1. (i)

We have $d_{u}(G)=d_{u}(G_{a.})$ unless $p\neq 2$ and $u\in\{2,6,10,14,18\}$. 2. (ii)

Suppose $p\neq 2$. Then $d_{2}(G)=d_{u}(G_{a.})+6$ and $d_{u}(G)=d_{u}(G_{a.})+2$ for $u\in\{6,10,14,18\}$. More precisely,

[TABLE]

Proof.

Recall that by [7] $d_{u}(G)\geq d_{u}(G_{a.})$, $d_{u}(G_{a})=7$ for $u\geq h=18$, and

[TABLE]

By Lemma 4.1 we can assume without loss of generality that $p\neq 2$ and $u\equiv 0\mod 2$, as otherwise $d_{u}(G)=d_{u}(G_{a.})$. Write $u=qv$ where $q$ and $v$ are coprime positive integers such that $q$ is a power of $p$ and $p$ does not divide $v$. Note that $q$ is odd and $v\equiv 0\mod 2$.

We first suppose that $u\leq h=18$.

Assume $u=2$. Then $q=1$ and $u=v=2$. By [1, Table 6], $d_{2}(G)=69.$ Let

[TABLE]

where $k\in K^{*}$ is an element of order 2. Then $y_{2}$ is a semisimple element of $G$ of order 2. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y_{2})^{0})=d_{1}(A_{1}D_{6})=69=d_{2}(G)$.

Assume $u=4$. Then $q=1$ and $u=v=4$. By [7] and [1, Table 6], $d_{4}(G)=d_{4}(G_{a})=33$. Let

[TABLE]

where $k\in K^{*}$ is an element of order 4. Then $y_{4}$ is a semisimple element of $G$ of order 4. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y_{4})^{0})=d_{1}(A_{3}A_{3}A_{1})=33=d_{4}(G)$.

Assume $u=6$. Then $q=1$ and $u=v=6$, or $q=p=3$ and $v=2$. Suppose first that $q=p=3$ and $v=2$. By [1, Table 6] a semisimple element $y$ of $G$ of order $2$ is such that $C_{G}(y)^{0}=A_{1}D_{6}$ or $E_{7}$. It follows from Proposition 2.2(i) that $d_{6}(G)=d_{3}(A_{1}D_{6})=23.$ Suppose now that $q=1$. By [1, Table 6], $d_{6}(G)=23.$ Let

[TABLE]

where $k\in K^{*}$ is an element of order 6. Then $y_{6}$ is a semisimple element of $G$ of order 6. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y_{6})^{0})=d_{1}(A_{3}A_{1}A_{1}T_{2})=23=d_{6}(G)$.

Assume $u=8$. Then $q=1$ and $u=v=8$. Let

[TABLE]

where $k\in K^{*}$ is an element of order 8. Then $y_{8}$ is a semisimple element of $G$ of order 8. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y_{8})^{0})=17$. Since, by [7], $d_{8}(G_{a})=17$, we deduce from Proposition 2.2(i) that $d_{8}(G)=17$.

Assume $u=10$. Then $q=1$ and $u=v=10$, or $q=p=5$ and $v=2$. Suppose first that $q=p=5$ and $v=2$. By [1, Table 6] a semisimple element $y$ of $G$ of order $2$ is such that $C_{G}(y)^{0}=A_{1}D_{6}$ or $E_{7}$. It follows from Proposition 2.2(i) that $d_{10}(G)=d_{5}(A_{1}D_{6})=15.$ Suppose now that $q=1$. By (4.1) and (4.2), a semisimple $y$ element of $G$ of order 10 is such that $C_{G}(y)^{0}$ is generated by at least four root subgroups $U_{\alpha}$ where $\alpha$ is a positive root of $G$. It follows from [7] that $d_{10}(G)>d_{10}(G_{a.})=13$. Hence by Lemma 2.6 $d_{10}(G)\geq 15$. Let

[TABLE]

where $k\in K^{*}$ is an element of order 10. Then $y_{10}$ is a semisimple element of $G$ of order 10. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence

[TABLE]

It now follows from Proposition 2.2(i) that $d_{10}(G)=15.$

Assume $u=12$. Then $q=1$ and $u=v=12$, or $q=p=3$ and $v=4$. Suppose first that $q=p=3$ and $v=4$. Consider the semisimple element $y_{4}$ of $G$ defined above. Then $C_{G}(y_{4})^{0}=A_{3}A_{3}A_{1}$ and $d_{q}(C_{G}(y_{4})^{0})=d_{3}(A_{3}A_{3}A_{1})=11$. Hence by Proposition 2.2(i), we obtain $d_{12}(G)\leq 11$. Suppose now that $q=1$ and $u=v=12$. Let

[TABLE]

where $k\in K^{*}$ is an element of order 12. Then $y_{12}$ is a semisimple element of $G$ of order 12. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y_{12})^{0})=11$.

Since, by [7], $d_{12}(G_{a})=11$, we deduce that $d_{12}(G)=11$ for all $p$.

Assume $u=14$. Then $q=1$ and $u=v=14$, or $q=p=7$ and $v=2$. Suppose first that $q=p=7$ and $v=2$. By [1, Table 6] a semisimple element $y$ of $G$ of order $2$ is such that $C_{G}(y)^{0}=A_{1}D_{6}$ or $E_{7}$. It follows from Proposition 2.2(i) that $d_{14}(G)=d_{7}(A_{1}D_{6})=11.$ Suppose now that $q=1$. By (4.1) and (4.2), a semisimple $y$ element of $G$ of order 14 is such that $C_{G}(y)^{0}$ is generated by at least two root subgroups $U_{\alpha}$ where $\alpha$ is a positive root of $G$. We deduce that $C_{G}(s)^{0}\neq A_{1}T_{6}$. It follows from [7] that $d_{14}(G)>d_{14}(G_{a.})=9$. Hence by Lemma 2.6 $d_{14}(G)\geq 11$. Let

[TABLE]

where $k\in K^{*}$ is an element of order 14. Then $y_{14}$ is a semisimple elements of $G$ of order 14. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence

[TABLE]

It now follows from Proposition 2.2(i) that $d_{14}(G)=11.$

Assume $u=16$. Then $q=1$ and $u=v=16$. Let

[TABLE]

where $k\in K^{*}$ is an element of order 16. Then $y_{16}$ is a semisimple element of $G$ of order 16. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y_{16})^{0})=9$. Since, by [7], $d_{16}(G_{a})=9$, we deduce from Proposition 2.2(i) that $d_{16}(G)=9$.

Assume $u=18$. Then $q=1$ and $u=v=18$, or $q=9$, $p=3$ and $v=2$. Suppose first that $q=9$, $p=3$ and $v=2$. By [1, Table 6] a semisimple element $y$ of $G$ of order $2$ is such that $C_{G}(y)^{0}=A_{1}D_{6}$ or $E_{7}$. It follows from Proposition 2.2(i) that $d_{18}(G)=d_{9}(A_{1}D_{6})=9$. Suppose now that $q=1$. By (4.1) and (4.2), a semisimple $y$ element of $G$ of order 18 is such that $C_{G}(y)^{0}$ is generated by at least one root subgroup $U_{\alpha}$ where $\alpha$ is a positive root of $G$. We deduce that $C_{G}(y)^{0}\neq T_{7}$. It follows from [7] that $d_{18}(G)>d_{18}(G_{a.})=7$. Hence by Lemma 2.6 $d_{18}(G)\geq 9$. Let

[TABLE]

where $k\in K^{*}$ is an element of order 18. Then $y_{18}$ is a semisimple element of $G$ of order 18. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence

[TABLE]

It now follows from Proposition 2.2(i) that $d_{18}(G)=9.$

We now suppose that $u>h=18$. Recall that $p\neq 2$ and $v\equiv 0\mod 2$. Assume $v>h$. Note that $v\geq 20$. Let

[TABLE]

where $k\in K^{*}$ is an element of order $v$. Then $y$ is a semisimple element of $G$ of order $v\geq 20$. Moreover by (4.1) and (4.2) we get

[TABLE]

Hence $d_{q}(C_{G}(y)^{0})=d_{q}(T_{7})=7$ and so by Proposition 2.2(i) $d_{u}(G)=7$.

We finally suppose that $u>h=18$ and $v\leq h$. Since $v\equiv 0\mod 2$ we have $v\in\{2,4,6,8,10,12,14,16,18\}$.

Assume $v=2$. As $u\geq 20$ we have $q\geq 10$. Now $y_{2}$ is a semisimple element of $G$ of order $2$ with $C_{G}(y_{2})^{0}=A_{1}D_{6}$. As $q\geq 10$, we have $d_{q}(C_{G}(y_{2})^{0})=d_{q}(A_{1}D_{6})=7$.

Assume $v=4$. As $u\geq 20$ we have $q\geq 5$. Now $y_{4}$ is a semisimple element of $G$ of order $4$ with $C_{G}(y_{4})^{0}=A_{3}A_{3}A_{1}$. As $q\geq 4$, we have $d_{q}(C_{G}(y_{4})^{0})=d_{q}(A_{3}A_{3}A_{1})=7$.

Assume $v=6$. As $u\geq 20$ and $p\neq 2$ we have $q\geq 5$. Now $y_{6}$ is a semisimple element of $G$ of order $6$ with $C_{G}(y_{6})^{0}=A_{3}A_{1}A_{1}T_{2}$. As $q\geq 4$, we have $d_{q}(C_{G}(y_{6})^{0})=d_{q}(A_{3}A_{1}A_{1}T_{2})=7$.

Assume $v=8$. As $u\geq 20$ we have $q\geq 3$. Now $y_{8}$ is a semisimple element of $G$ of order $8$ with $C_{G}(y_{8})^{0}=A_{2}A_{1}A_{1}T_{3}$. As $q\geq 3$, we have $d_{q}(C_{G}(y_{8})^{0})=d_{q}(A_{2}A_{1}A_{1}T_{3})=7$.

Assume $v=10$. As $u\geq 20$ and $p\neq 2$ we have $q\geq 3$. Now $y_{10}$ is a semisimple element of $G$ of order $10$ with $C_{G}(y_{10})^{0}=A_{2}A_{1}T_{4}$. As $q\geq 3$, we have $d_{q}(C_{G}(y_{10})^{0})=d_{q}(A_{2}A_{1}T_{4})=7$.

Assume $v\in\{12,14\}$. As $u\geq 20$ and $p\neq 2$ we have $q\geq 3$. Now $y_{12}$ and $y_{14}$ are semisimple elements of $G$ of respective orders $12$ and $14$ with $C_{G}(y_{12})^{0}=A_{1}A_{1}T_{5}$ and $C_{G}(y_{14})^{0}=A_{1}A_{1}T_{5}$ . As $q\geq 2$, we have $d_{q}(C_{G}(y_{12})^{0})=d_{q}(C_{G}(y_{14})^{0})=d_{q}(A_{1}A_{1}T_{5})=7$.

Assume $v\in\{16,18\}$. As $u\geq 20$ and $p\neq 2$ we have $q\geq 3$. Now $y_{16}$ and $y_{18}$ are semisimple elements of $G$ of respective orders $16$ and $18$ with $C_{G}(y_{16})^{0}=A_{1}T_{6}$ and $C_{G}(y_{18})^{0}=A_{1}T_{6}$. As $q\geq 2$, we have $d_{q}(C_{G}(y_{16})^{0})=d_{q}(C_{G}(y_{18})^{0})=d_{q}(A_{1}T_{6})=7$.

Hence in the case where $u>h$ and $v\leq h$, by Proposition 2.2(i), we get $d_{u}(G)=7.$ ∎

5 Some properties of ${\rm Spin}_{n}(K)$

Let $K$ now denote an algebraically closed field $K$ of characteristic $p\neq 2$. We give a characterization of semisimple elements of ${\rm Spin}_{n}(K)$ of a given order (see Lemma 5.1 below). Before doing so we recall some properties of the group ${\rm Spin}_{n}(K)$ and the canonical surjective map ${\rm Spin}_{n}(K)\rightarrow{\rm SO}_{n}(K)$, where $n\geq 7$ is a positive integer. Write $n=2\ell$ or $n=2\ell+1$ for some integer $\ell\geq 1$, according respectively as $n$ is even or odd.

Let $V$ be the natural module for ${\rm SO}_{n}(K)$ and let

[TABLE]

(according respectively as $n$ is even or odd) be a standard basis of $V$. That is, $(e_{i},f_{i})=1$, $(e_{i},e_{i})=(f_{i},f_{i})=0$ for $1\leq i\leq\ell$, $(e_{i},f_{j})=(e_{i},e_{j})=(f_{i},f_{j})=0$ for $i\neq j$ such that $1\leq i,j\leq\ell$ and if $n$ is odd then $(d,d)=1$ and $(d,e_{i})=(d,f_{i})$ for $1\leq i\leq\ell$. Here $(,):V\times V\rightarrow K$ denotes the non-degenerate symmetric bilinear form associated to $V$.

Let $T_{0}(V,K)$ be the $K$-algebra of polynomials in $e_{1},f_{1},\dots,e_{\ell},f_{\ell}$ if $n$ is even (respectively, in $e_{1},f_{1},\dots,e_{\ell},f_{\ell},d$ if $n$ is odd). Also let $I(V,K)$ be the ideal of $T_{0}(V,K)$ generated by $\{vv-(v,v):v\in V\}$ and let $C_{0}(V,K)=T_{0}(V,K)/I(V,K)$.

Let ${}^{*}:C_{0}(V,K)\rightarrow C_{0}(V,K)$ be the $K$-linear map such that for any even positive integer $r$ and elements $x_{1},\dots,x_{r}\in\mathcal{B}$, we have the equality $(x_{1}\dots x_{r})^{*}=x_{r}\dots x_{1}$ (in $C_{0}(V,K))$. Note that for $x$ in $V$, we have $x^{*}x=xx^{*}=xx=(x,x)$.

We can now define the spin group ${\rm Spin}_{n}(K)$ as an abstract group, namely ${\rm Spin}_{n}(K)$ consists of the elements $t$ of $C_{0}(V,K)$ such that:

[TABLE]

and the map

[TABLE]

has determinant 1.

Given an element $t$ of ${\rm Spin}_{n}(K)$, consider the map $\phi_{t}:V\rightarrow V$ defined by $\phi_{t}(x)=txt^{-1}$ for $x$ in $V$. Then for any $x$ in $V$,

[TABLE]

and so $\phi_{t}$ is an element of ${\rm SO}_{n}(K)$. In fact the map $\phi$:

[TABLE]

is a surjective homomorphism with kernel equal to $\{1,-1\}$.

We can now characterize a preimage of a semisimple element of ${\rm SO}_{n}(K)$ of a given order under the canonical surjective map ${\rm Spin}_{n}(K)\rightarrow{\rm SO}_{n}(K)$, where $n\geq 7$ is a positive integer and $K$ is an algebraically closed field of characteristic $p\neq 2$.

Lemma 5.1**.**

Let $g$ be an element of ${\rm SO}_{n}(K)$ defined over an algebraically closed field $K$ of characteristic $p\neq 2$. Suppose $g$ is a semisimple element of ${\rm SO}_{n}(K)$ of order $u$, and let $\omega$ be a $u$-th root of unity of $K$. Let $\pm w$ be any of the two preimages of $g$ under the canonical surjective map $\phi:{\rm Spin}_{n}(K)\rightarrow{\rm SO}_{n}(K)$. The following assertions hold.

1. (i)

The order of $w$ in ${\rm Spin}_{n}(K)$ is divisible by $u$. 2. (ii)

There exits $t\in\{\pm w\}$ of order $u$ if and only if $u$ is odd, or $u$ is even and the number of eigenvalues of $\phi(t)$ of the form $\omega^{i}$ with $i$ odd is divisible by $4$. 3. (iii)

If neither $w$ nor $-w$ has order $u$, then $u$ is even and $\pm w$ has order $2u$.

Proof.

This is a classical result which follows from the properties of the canonical surjective map $\phi:{\rm Spin}_{n}(K)\rightarrow{\rm SO}_{n}(K)$ described above. We give a sketch of the argument.

We first consider part (i). Let ${g}$ be any semisimple element of ${\rm SO}_{n}(K)$ of order $u$. Let $w$ be a preimage of $g$ under $\phi$ (so that $-w$ is the other preimage of $g$ under $\phi$) and let $m$ be the order of $w$. We have $g^{m}=\phi(w)^{m}=\phi(w^{m})=\phi(1)=1$. Hence $u$, which is the order of $g$, divides the order of a preimage of $g$ under $\phi$. This establishes part (i).

We now consider parts (ii) and (iii). Let $V$ be the natural module for ${\rm SO}_{n}(K)$. There is a standard basis $\mathcal{B}$ of $V$, where $\mathcal{B}=(e_{1},f_{1},\dots,e_{\ell},f_{\ell})$ or $(e_{1},f_{1},\dots,e_{\ell},f_{\ell},d)$ according respectively as $n$ is even or odd, such that the matrix of ${g}$ with respect to $\mathcal{B}$ is diagonal with entries $d_{i}$ satisfying $d_{i}^{u}=1$ for $1\leq i\leq n$. Note that if $1\leq j\leq\ell$ then $d_{2j-1}=d_{2j}^{-1}$ is a power of $\omega$, and if $n$ is odd then $d_{n}=1$. Moreover without loss of generality $d_{1}=\omega$ and $d_{2}=\omega^{-1}$. For $1\leq j\leq\ell$, let ${g_{j}}$ be the element of ${\rm SO}_{n}(K)$ such that, with respect to $\mathcal{B}$, ${g_{j}}$ is diagonal and $({g_{j}})_{2j-1,2j-1}=d_{j}$, $({g_{j}})_{2j,2j}=d_{j+1}$, $({g_{j}})_{k,k}=1$ for $k\not\in\{2j-1,2j\}$. Note that ${g}={g_{1}}\dots g_{\ell}$. In the decomposition ${g}={g_{1}}\dots g_{\ell}$ delete the elements ${g_{j}}$ with $1\leq i\leq\ell$ such that ${g_{j}}=1$, and renumbering the elements ${g_{j}}$ where $1\leq j\leq\ell$, if necessary, write ${g}={g_{1}}\dots g_{\iota}$ where $1\leq\iota\leq\ell$ and ${g_{1}},\dots,{g_{\iota}}$ are not the identity element.

Let $1\leq j\leq\iota$ and let $1\leq k\leq\iota$ be such that $({g_{j}})_{2k-1,2k-1}$ and $({g_{j}})_{2k,2k}$ are both not 1. Write $\theta_{j}=({g_{j}})_{2k-1,2k-1}$. Note that $\theta_{j}\neq 1$ is a power of $\omega$ and $\theta_{j}^{-1}=({g_{j}})_{2k,2k}$. Moreover ${g_{j}}(e_{k})=\theta_{j}e_{k}$, ${g_{j}}(f_{k})=\theta_{j}^{-1}f_{k}$ and ${g_{j}}$ fixes every other element of $\mathcal{B}$. Write $\theta_{j}=\omega^{l_{j}}$ for some positive integer $l_{j}$.

For a nonsingular element $x$ in $V$, let $R_{x}$ be the reflection:

[TABLE]

An easy check yields

[TABLE]

Let

[TABLE]

Then $x_{j}\in C_{0}(V,K)$ and $x_{j}x_{j}^{*}=16(1-\theta_{j})^{2}\theta_{j}^{-1}$. Let $\tau_{j}\in K$ be such that $\tau_{j}^{2}=\theta_{j}^{-1}=\omega^{-l_{j}}$, furthermore if $u$ and $l_{j}$ are both odd take $\tau_{j}=\omega^{(u-l_{j})/2}$, and if $l_{j}$ is even take $\tau_{j}=\omega^{-l_{j}/2}$. Let

[TABLE]

Then $t_{j}$ is an element of ${\rm Spin}_{n}(K)$ and the image of $t_{j}$ under the canonical surjective map $\phi:{\rm Spin}_{n}(K)\rightarrow{\rm SO}_{n}(K)$ is ${g_{j}}$. Furthermore, for any positive integer $m$,

[TABLE]

and so

[TABLE]

Finally, let $t=t_{1}\dots t_{\iota}$. Then $t\in{\rm Spin}_{n}(K)$ and $\phi(t)={g}={g_{1}}\dots g_{\iota}$. Now for $1\leq i,j\leq\iota$, $t_{i}$ and $t_{j}$ commute and so

[TABLE]

It follows from (5.1) that $t^{u}=1$ if and only if $u$ is odd or $u$ is even and the number of eigenvalues of $\phi(t)=g$ of the form $\omega^{i}$ with $i$ odd is divisible by $4$. Otherwise $t^{u}=-1$ and $t^{2u}=1$. Note that the other preimage of $g$ under $\phi$ is $-t$ and $(-t)^{u}=t^{u}$ if $u$ is even, otherwise $(-t)^{u}=-t^{u}$. Parts (ii) and (iii) now follow from part (i). ∎

6 Upper bounds for $d_{u}(G)$ for $G$ of classical type

Let $G$ be a simple algebraic group of classical type defined over an algebraically closed field $K$ of characteristic $p$. Let $u$ be a positive integer. In this section, we give an upper bound for $d_{u}(G)$. Since by Proposition 3 $d_{u}(G)\leq d_{u}(G_{s.c.})$ we assume without loss of generality that $G$ is of simply connected type.

Unless otherwise stated, we use the notation introduced at the end of §1. Recall, we let $h$ be the Coxeter number of $G$ and write:

$u=qv$ where $q$ and $v$ are coprime positive integers such that $q$ is a power of $p$ and $p$ does not divide $v$,

$h=zu+e$ where $z,e$ are nonnegative integers such that $0\leq e\leq q-1$,

$h=\alpha v+\beta$ where $\alpha,\beta$ are nonnegative integers such that $0\leq\alpha\leq v-1$,

$\alpha=\gamma q+\delta$ where $\gamma,\delta$ are nonnegative integers such that $0\leq\delta\leq q-1$.

An easy check yields $\gamma=z$. Also if $e=0$ then $\beta=\delta=0$. Furthermore, if $\beta=0$ then $\delta=0$ if and only if $e=0$.

Finally for an nonnegative integer $r$, we let $\epsilon_{r}\in\{0,1\}$ be [math] if $r$ is even, otherwise $\epsilon_{r}=1$, and set $\sigma_{r}\in\{0,1\}$ to be 1 if $r=0$, otherwise $\sigma_{r}=0$.

Moreover, we also let $\omega$ be a $v$-th root of 1 in $K$ and consider the following diagonal blocks $M_{1}$, $M_{2}$, $M_{3}$, $M_{4}$, $M_{5}$, ${M_{6}}$, $M_{7}$, $M_{8}$ and $M_{9}$ where:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and if $\beta\geq 2$ is even then

[TABLE]

[TABLE]

and if $v\geq 4$ is even then

[TABLE]

[TABLE]

[TABLE]

We also write $(1)$ for $I_{1}={\rm diag}(1)$ and $(-1)$ for $-I_{1}={\rm diag}(-1)$. More generally given a positive integer $i$, we write $(\omega^{i})$ for ${\rm diag}(\omega^{i})$ Finally, given some nonnegative integers $r_{1},\dots,r_{11}$ and some nonnegative integers $s_{1},\dots,s_{v-1}$we denote by

[TABLE]

the diagonal matrix consisting of $r_{i}$ blocks $M_{i}$ for $1\leq i\leq 9$, $I_{r_{10}}$, $-I_{r_{11}}$ and $\omega^{j}I_{s_{j}}$ for $1\leq j\leq v-1$.

6.1 $G$ is of type $A_{\ell}$

Let $G=(A_{\ell})_{s.c.}$ with $\ell\geq 1$ defined over an algebraically closed field $K$ of characteristic $p$. Here $|\Phi|=\ell(\ell+1)$ and $h=\ell+1$.

Lemma 6.1**.**

Let $G=(A_{\ell})_{s.c.}$ be defined over an algebraically closed field of characteristic $p$. Let $u=qv$ be a positive integer where $q$ and $v$ are coprime positive integers such that $q$ is a power of $p$ and $p$ does not divide $v$. Write $h=zu+e=\alpha v+\beta$ and $\alpha=zq+\delta$ where $z$, $e$, $\alpha$, $\beta$, $\delta$ are nonnegative integers such that $e<u$, $\beta<v$, and $\delta<q$. Let $y$ be the element of $G_{s.c.}$ of order $v$ defined in Table 4 (see §11). Then $C_{G}(y)^{0}$ and $d_{q}(C_{G}(y)^{0})$ are given in Table 4 and $d_{q}(C_{G}(y)^{0})$ is an upper bound for $d_{u}(G)$.

Proof.

We first determine $C_{G}(y)^{0}$. Suppose that $\epsilon_{v}=1$ or $(\epsilon_{v},\epsilon_{\alpha})=(0,0)$. As any nontrivial eigenvalue of $y$ can be paired with its inverse, $y$ is an element of $G$ of order $v$. Also $y$ has $\alpha+\epsilon_{\beta}$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\lfloor\beta/2\rfloor$, and every other eigenvalue of $y$ occurs with multiplicity $\alpha$. Therefore $C_{G}(y)^{0}=A_{\alpha}^{\beta}A_{\alpha-1}^{v-\beta}T_{v-1}$.

Suppose that $(\epsilon_{v},\epsilon_{\alpha},\epsilon_{\beta})=(0,1,1)$. As any nontrivial eigenvalue of $y$ can be paired with its inverse, $y$ is an element of $G$ of order $v$. Also $y$ has $\alpha+1$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq(\beta-1)/2$, and every other eigenvalue of $y$ occurs with multiplicity $\alpha$. Therefore $C_{G}(y)^{0}=A_{\alpha}^{\beta}A_{\alpha-1}^{v-\beta}T_{v-1}$.

Suppose that $(\epsilon_{v},\epsilon_{\alpha},\epsilon_{\beta})=(0,1,0)$ and $\beta\geq 2$. As any eigenvalue of $y$ can be paired with its inverse, $y$ is an element of $G$ of order $v$. Also $y$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq(\beta-2)/2$, and every other eigenvalue of $y$ occurs with multiplicity $\alpha$. Therefore $C_{G}(y)^{0}=A_{\alpha}^{\beta}A_{\alpha-1}^{v-\beta}T_{v-1}$.

Suppose finally that $(\epsilon_{v},\epsilon_{\alpha})=(0,1)$ and $\beta=0$. As any eigenvalue of $y$ can be paired with its inverse, $y$ is an element of $G$ of order $v$. Also $y$ has $\alpha-1$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, and every other eigenvalue of $y$ occurs with multiplicity $\alpha$. Therefore $C_{G}(y)^{0}=A_{\alpha}A_{\alpha-1}^{v-2}A_{\alpha-2}T_{v-1}$.

Note that by Proposition 2.2(i), $d_{u}(G)\leq d_{q}(C_{G}(y)^{0})$ and so $d_{q}(C_{G}(y)^{0})$ is an upper bound for $d_{u}(G)$. It remains to calculate $d_{q}(C_{G}(y)^{0})$.

By Lemma 2.4,

[TABLE]

[TABLE]

and

[TABLE]

where for determining $d_{q}(A_{\alpha-2})$ we do the Euclidean division of $\alpha-1$ by $q$:

[TABLE]

Assume first that $(\epsilon_{v},\epsilon_{\alpha})\neq(0,1)$ or $\beta>0$. We then have

[TABLE]

By Proposition 2.2(iii), (6.1) and (6.2)

[TABLE]

Assume now that $(\epsilon_{v},\epsilon_{\alpha})=(0,1)$ and $\beta=0$. Since $\beta=0$, recall that $e=0$ if and only if $\delta=0$. We then have

[TABLE]

By Proposition 2.2(iii), (6.1), (6.2) and (6.3)

[TABLE]

∎

6.2 $G$ is of type $C_{\ell}$

Let $G=(C_{\ell})_{{\rm s.c}}$ be a simply connected group of type ${C}_{\ell}$ with $\ell\geq 2$ defined over an algebraically closed field $K$ of characteristic $p$. Here $|\Phi|=2\ell^{2}$ and $h=2\ell$.

Lemma 6.2**.**

Let $G=(C_{\ell})_{{\rm s.c}}$ be defined over an algebraically closed field of characteristic $p$. Let $u=qv$ be a positive integer where $q$ and $v$ are positive integers such that $q$ is a power of $p$. Write $h=zu+e=\alpha v+\beta$ and $\alpha=zq+\delta$ where $z$, $e$, $\alpha$, $\beta$, $\delta$ are nonnegative integers such that $e<u$, $\beta<v$, and $\delta<q$. Let $y$ be the element of $G$ of order $v$ defined in Table 5 (see §11). Then $C_{G}(y)^{0}$ and $d_{q}(C_{G}(y)^{0})$ are given in Table 5 and $d_{q}(C_{G}(y)^{0})$ is an upper bound for $d_{u}(G)$.

Proof.

Since $h=2\ell$, note that if $\epsilon_{v}=1$ then $\epsilon_{\alpha}=\epsilon_{\beta}$ and $\epsilon_{u}=\epsilon_{q}$, whereas if $\epsilon_{v}=0$ then $\epsilon_{u}=0$, $\epsilon_{q}=1$ and $\epsilon_{\beta}=0$. We first determine $C_{G}(y)^{0}$. Suppose that $\epsilon_{v}=1$. As any eigenvalue of $y$ can be paired with its inverse, $y$ is an element of $G$ of order $v$. Also $y$ has $\alpha+\epsilon_{\beta}$ eigenvalues equal to $1$, [math] eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\lfloor\beta/2\rfloor$, and every other eigenvalue of $y$ occurs with multiplicity $\alpha$. Therefore $C_{G}(y)^{0}=A_{\alpha}^{\lfloor\frac{\beta}{2}\rfloor}A_{\alpha-1}^{\frac{v-1}{2}-\lfloor\frac{\beta}{2}\rfloor}C_{\lceil\frac{\alpha}{2}\rceil}T_{\frac{v-1}{2}}$.

Suppose that $(\epsilon_{v},\epsilon_{\alpha})=(0,0)$. As any eigenvalue of $y$ can be paired with its inverse, $y$ is an element of $G$ of order $v$. Also $y$ has $\alpha$ eigenvalues equal to $1$, $\alpha$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\beta/2$, and every other eigenvalue of $y$ occurs with multiplicity $\alpha$. Therefore $C_{G}(y)^{0}=A_{\alpha}^{\frac{\beta}{2}}A_{\alpha-1}^{\frac{v}{2}-1-\frac{\beta}{2}}C_{\frac{\alpha}{2}}^{2}T_{\frac{v}{2}-1}$.

Suppose that $(\epsilon_{v},\epsilon_{\alpha})=(0,1)$ and $\beta\geq 2$. As any eigenvalue of $y$ can be paired with its inverse, $y$ is an element of $G$ of order $v$. Also $y$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq(\beta-2)/2$, and every other eigenvalue of $y$ occurs with multiplicity $\alpha$. Therefore $C_{G}(y)^{0}=A_{\alpha}^{\frac{\beta}{2}-1}A_{\alpha-1}^{\frac{v}{2}-\frac{\beta}{2}}C_{\frac{\alpha+1}{2}}^{2}T_{\frac{v}{2}-1}$.

Suppose finally that $(\epsilon_{v},\epsilon_{\alpha})=(0,1)$ and $\beta=0$. As any eigenvalue of $y$ can be paired with its inverse, $y$ is an element of $G$ of order $v$. Also $y$ has $\alpha-1$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, and every other eigenvalue of $y$ occurs with multiplicity $\alpha$. Therefore $C_{G}(y)^{0}=A_{\alpha-1}^{\frac{v}{2}-1}C_{\frac{\alpha+1}{2}}C_{\frac{\alpha-1}{2}}T_{\frac{v}{2}-1}$.

By Proposition 2.2(i), $d_{u}(G)\leq d_{q}(C_{G}(y)^{0})$ and so $d_{q}(C_{G}(y)^{0})$ is an upper bound for $d_{u}(G)$. It remains to calculate $d_{q}(C_{G}(y)^{0})$.

Note that by Lemma 2.4,

[TABLE]

[TABLE]

[TABLE]

and if $\alpha$ is odd then

[TABLE]

Assume first that $\epsilon_{v}=1$. Recall that $\epsilon_{u}=\epsilon_{q}$ and $\epsilon_{\alpha}=\epsilon_{\beta}$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that $(\epsilon_{v},\epsilon_{\alpha})=(0,0)$. Recall that $\epsilon_{q}=1$ and $\epsilon_{\beta}=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that $(\epsilon_{v},\epsilon_{\alpha})=(0,1)$ and $\beta\geq 2$. Recall that $\epsilon_{q}=1$ and $\epsilon_{\beta}=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume finally that $(\epsilon_{v},\epsilon_{\alpha})=(0,1)$ and $\beta=0$. Recall that $\epsilon_{q}=1$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

∎

6.3 $G$ is of type $B_{\ell}$

Let $G=(B_{\ell})_{{\rm s.c}}$ be a simply connected group of type ${B}_{\ell}$ with $\ell\geq 3$ defined over an algebraically closed field $K$ of characteristic $p$. Here $|\Phi|=2\ell^{2}$, $h=2\ell$ and $G={\rm Spin}_{h+1}(K)$ if $p\neq 2$, otherwise $G={\rm SO}_{h+1}(K)$. Given an element $g$ in $G$, we let $\overline{g}$ be the image of $g$ under the canonical surjective map $G\rightarrow{\rm SO}_{h+1}(K)$.

Lemma 6.3**.**

*Let $G=(B_{\ell})_{{\rm s.c}}$ where $\ell\geq 3$ be defined over an algebraically closed field $K$ of characteristic $p$. Let $u=qv$ be a positive integer where $q$ and $v$ are positive integers such that $q$ is a power of $p$ and $p$ does not divide $v$. Write $h=zu+e=\alpha v+\beta$ and $\alpha=zq+\delta$ where $z$, $e$, $\alpha$, $\beta$, $\delta$ are nonnegative integers such that $e<u$, $\beta<v$, and $\delta<q$. Consider the following conditions:

1. (a)

$\epsilon_{v}=1$. 2. (b)

$v=2$. 3. (c)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$, $v>2$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)\in\{(0,0,0),(0,2,3),(2,0,3),(2,2,2)\}.$ 4. (d)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$* and $(v\mod 4,\beta\mod 4,\ell\mod 4)\in\{(0,2,1),(2,2,0)\}.$* 5. (e)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$, $\beta>0$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)\in\{(0,0,2),(2,0,1)\}.$ 6. (f)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$, $v>2$, $\beta=0$ and $(v\mod 4,\ell\mod 4)\in\{(0,2),(2,1)\}.$ 7. (g)

$(\epsilon_{v},\epsilon_{\alpha})=(0,0)$, $v>2$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)\in\{(0,0,0),(0,2,3),(2,0,0),(2,2,3)\}.$ 8. (h)

$(\epsilon_{v},\epsilon_{\alpha})=(0,0)$* and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(2,2,1)$.* 9. (i)

$(\epsilon_{v},\epsilon_{\alpha})=(0,0)$, $\beta>0$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)\in\{(0,2,1),(2,0,2)\}.$ 10. (j)

$(\epsilon_{v},\epsilon_{\alpha})=(0,0)$, $\beta>0$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)\in(0,0,2)$. 11. (k)

$(\epsilon_{v},\epsilon_{\alpha})=(0,0)$, $v>2$, $\beta=0$ and $(v\mod 4,\ell\mod 4)=(2,2)$.

Let $\overline{y}$ be the element of ${\rm SO}_{h+1}(K)$ of order $v$ defined in Table 6 (see §11). Then $\overline{y}$ has a preimage $y$ in $G$ of order $v$, and $C_{{\rm SO}_{h+1}}(\overline{y})^{0}$ and $d_{q}(C_{G}(y)^{0})=d_{q}(C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0})$ are given in Table 6. Furthermore $d_{q}(C_{G}(y)^{0})$ is an upper bound for $d_{u}(G)$.

Proof.

Note that by Lemma 2.3(iii), $d_{q}(C_{G}(y)^{0})=d_{q}(C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0})$. Since $h=2\ell$, note that if $\epsilon_{v}=1$ then $\epsilon_{\alpha}=\epsilon_{\beta}$ and $\epsilon_{u}=\epsilon_{q}$, whereas if $\epsilon_{v}=0$ then $\epsilon_{u}=0$, $\epsilon_{q}=1$ and $\epsilon_{\beta}=0$. Note also that if $\ell\equiv 2\mod 4$, $v\equiv 0\mod 4$ and $\beta=0$ then $\alpha$ is odd, as otherwise on one hand we would have $2\ell\equiv 4\mod 8$ and on the other $2\ell=\alpha v\equiv 0\mod 8$, a contradiction. We first determine $C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0}$. Suppose that case (a) holds so that $v$ is odd. As any eigenvalue of $\overline{y}$ different from 1 can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+1}(K)$ of order $v$. Furthermore, since $v$ is odd, by Lemma 5.1 it follows that $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1+\epsilon_{\beta}$ eigenvalues equal to $1$, [math] eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\lfloor\beta/2\rfloor$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0}=A_{\alpha}^{\lfloor\frac{\beta}{2}\rfloor}A_{\alpha-1}^{\frac{v-1}{2}-\lfloor\frac{\beta}{2}\rfloor}B_{\lceil\frac{\alpha}{2}\rceil}T_{\frac{v-1}{2}}$.

Suppose that case (b) holds so that $v=2$. As any eigenvalue of $\overline{y}$ different from 1 can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+1}(K)$ of order $v$. Now the number $N$ of eigenvalues $\omega^{i}=(-1)^{i}$ of $\overline{y}$ with $i$ odd is equal to $\ell$, $\ell-1$, $\ell+2$ or $\ell+1$ according respectively as $\ell\mod 4$ is 0, 1, 2 or 3. Since $N$ is divisible by 4, it follows from Lemma 5.1 that $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also

[TABLE]

Suppose that case (c) holds. As any eigenvalue of $\overline{y}$ different from 1 can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+1}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (c), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\beta/2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}}A_{\alpha-1}^{\frac{v}{2}-1-\frac{\beta}{2}}B_{\frac{\alpha-1}{2}}D_{\frac{\alpha+1}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (d) or (e) holds. As any eigenvalue of $\overline{y}$ different from 1 can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+1}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (d) or (e), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+2$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where if case (d) holds then $i$ is any integer with $1\leq|i|\leq(\beta/2-1)$ and if case $(e)$ holds then $i$ any integer with $2\leq|i|\leq\beta/2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}-1}A_{\alpha-1}^{\frac{v}{2}-\frac{\beta}{2}}B_{\frac{\alpha+1}{2}}D_{\frac{\alpha+1}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (f) holds. As any eigenvalue of $\overline{y}$ different from 1 can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+1}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (f), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+2$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, $\alpha$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $2\leq|i|\leq v/2-1$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha-1$. Therefore $C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0}=A_{\alpha-1}^{\frac{v}{2}-2}A_{\alpha-2}B_{\frac{\alpha+1}{2}}D_{\frac{\alpha+1}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (g) holds. As any eigenvalue of $\overline{y}$ different from 1 can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+1}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (g), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\beta/2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}}A_{\alpha-1}^{\frac{v}{2}-1-\frac{\beta}{2}}B_{\frac{\alpha}{2}}D_{\frac{\alpha}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (h) holds. As any eigenvalue of $\overline{y}$ different from 1 can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+1}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (h), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $2\leq|i|\leq\beta/2$, $(\alpha+1)$ eigenvalues equal to $\omega^{\frac{v}{2}-1}$, $(\alpha+1)$ eigenvalues equal to $\omega^{-\left(\frac{v}{2}-1\right)}$ and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}}A_{\alpha-1}^{\frac{v}{2}-\frac{\beta}{2}-1}B_{\frac{\alpha}{2}}D_{\frac{\alpha}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (i) or (j) holds. As any eigenvalue of $\overline{y}$ different from 1 can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+1}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (i) or (j), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha+2$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where if case (i) holds then $i$ is any integer with $1\leq|i|\leq(\beta/2-1)$ and if case $(j)$ holds then $i$ any integer with $2\leq|i|\leq\beta/2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}-1}A_{\alpha-1}^{\frac{v}{2}-\frac{\beta}{2}}B_{\frac{\alpha}{2}}D_{\frac{\alpha+2}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (k) holds. As any eigenvalue of $\overline{y}$ different from 1 can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+1}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumption of case (k), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha+2$ eigenvalues equal to $-1$, $\alpha$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq v/2-2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha-1$. Therefore $C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0}=A_{\alpha-1}^{\frac{v}{2}-2}A_{\alpha-2}B_{\frac{\alpha}{2}}D_{\frac{\alpha+2}{2}}T_{\frac{v}{2}-1}$.

By Proposition 2.2(i), $d_{u}(G)\leq d_{q}(C_{G}(y)^{0})$ and so $d_{q}(C_{G}(y)^{0})$ is an upper bound for $d_{u}(G)$. It remains to calculate $d_{q}(C_{{\rm SO}_{h+1}}(\overline{y})^{0})=d_{q}(C_{G}(y)^{0})$.

It follows from Lemma 2.4 that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

if $\alpha$ is odd then

[TABLE]

[TABLE]

and if $\alpha$ is even then

[TABLE]

and

[TABLE]

Assume first that case (a) holds so that $\epsilon_{v}=1$. Recall that $\epsilon_{u}=\epsilon_{q}$ and $\epsilon_{\alpha}=\epsilon_{\beta}$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (b) holds so that $v=2$, $\epsilon_{u}=\epsilon_{v}=0$ and $\epsilon_{q}=1$. Recall that $\alpha=\ell$ and $\beta=0$. Note also that $e=0$ if and only if $\delta=0$. We have

[TABLE]

If $\alpha\equiv 0\mod 4$ then $z$ is even if $\delta=0$, and by Proposition 2.2(i)

[TABLE]

If $\alpha\equiv 1\mod 4$ then by Proposition 2.2(iii)

[TABLE]

If $\alpha\equiv 2\mod 4$ then $z$ is even if $\delta=0$, and by Proposition 2.2(iii)

[TABLE]

If $\alpha\equiv 3\mod 4$ then by Proposition 2.2(iii)

[TABLE]

Assume that case (c) holds. Recall that $\epsilon_{q}=\epsilon_{\alpha}=1$ and $\epsilon_{u}=\epsilon_{v}=\epsilon_{\beta}=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (d) or case (e) holds. Recall that $\epsilon_{q}=\epsilon_{\alpha}=1$ and $\epsilon_{u}=\epsilon_{v}=\epsilon_{\beta}=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (f) holds. Recall that $\epsilon_{q}=\epsilon_{\alpha}=1$, $\epsilon_{u}=\epsilon_{v}=\epsilon_{\beta}=0$ and $\beta=0$. Also $\delta=0$ if and only if $e=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (g) or case (h) holds. Recall that $\epsilon_{q}=1$ and $\epsilon_{u}=\epsilon_{v}=\epsilon_{\alpha}=\epsilon_{\beta}=0$. Also if $\delta=0$ then $z$ must be even. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (i) or case (j) holds. Recall that $\epsilon_{q}=1$ and $\epsilon_{u}=\epsilon_{v}=\epsilon_{\alpha}=\epsilon_{\beta}=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (k) holds. Recall that $\epsilon_{q}=1$, $\epsilon_{u}=\epsilon_{v}=\epsilon_{\alpha}=\epsilon_{\beta}=0$ and $\beta=0$. Also $\delta=0$ if and only if $e=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

∎

6.4 $G$ is of type $D_{\ell}$

Let $G=(D_{\ell})_{{\rm s.c}}$ be a simply connected group of type ${D}_{\ell}$ with $\ell\geq 4$ defined over an algebraically closed field $K$ of characteristic $p$. Here $|\Phi|=2\ell(\ell-1)$, $h=2\ell-2$ and $G={\rm Spin}_{h+2}(K)$ if $p\neq 2$, otherwise $G={\rm SO}_{h+2}(K)$. Given an element $g$ in $G$, we let $\overline{g}$ be the image of $g$ under the canonical surjective map $G\rightarrow{\rm SO}_{h+2}(K)$.

Lemma 6.4**.**

*Let $G=(D_{\ell})_{{\rm s.c}}$ where $\ell\geq 4$ be defined over an algebraically closed field $K$ of characteristic $p$. Let $u=qv$ be a positive integer where $q$ and $v$ are positive integers such that $q$ is a power of $p$ and $p$ does not divide $v$. Write $h=zu+e=\alpha v+\beta$ and $\alpha=zq+\delta$ where $z$, $e$, $\alpha$, $\beta$, $\delta$ are nonnegative integers such that $e<u$, $\beta<v$, and $\delta<q$. Consider the following conditions:

1. (a)

$\epsilon_{v}=1$. 2. (b)

$v=2$. 3. (c)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$, $v>2$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)\in\{(0,0,1),(0,2,0),(2,0,0),(2,2,3)\}.$ 4. (d)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$* and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(0,2,2).$* 5. (e)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$, $v>2$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(2,2,1).$ 6. (f)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$, $\beta>0$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(0,0,3).$ 7. (g)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$, $v>2$, $\beta>0$, $\beta\neq v-2$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(2,0,2).$ 8. (h)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$, $v>2$, $\beta=v-2$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(2,0,2).$ 9. (i)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$, $\beta=0$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(0,0,3).$ 10. (j)

$(\epsilon_{v},\epsilon_{\alpha})=(0,1)$, $v>2$, $\beta=0$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(2,0,2).$ 11. (k)

$(\epsilon_{v},\epsilon_{\alpha})=(0,0)$, $v>2$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)\in\{(0,0,1),(0,2,0),(2,0,1),(2,2,0)\}.$ 12. (l)

$(\epsilon_{v},\epsilon_{\alpha})=(0,0)$* and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(2,2,2)$.* 13. (m)

$(\epsilon_{v},\epsilon_{\alpha})=(0,0)$* and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(0,0,3)$.* 14. (n)

$(\epsilon_{v},\epsilon_{\alpha})=(0,0)$* and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(0,2,2)$.* 15. (o)

$(\epsilon_{v},\epsilon_{\alpha})=(0,0)$, $v>2$, $\beta>0$ and $(v\mod 4,\beta\mod 4,\ell\mod 4)=(2,0,3)$. 16. (p)

$(\epsilon_{v},\epsilon_{\alpha})=(0,0)$, $v>2$, $\beta=0$ and $(v\mod 4,\ell\mod 4)=(2,3)$.

Let $\overline{y}$ be the element of ${\rm SO}_{h+2}(K)$ of order $v$ defined in Table 7 (see §11). Then $\overline{y}$ has a preimage $y$ in $G$ of order $v$, and $C_{{\rm SO}_{h+2}}(\overline{y})^{0}$ and $d_{q}(C_{G}(y)^{0})=d_{q}(C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0})$ are given in Table 7. Furthermore $d_{q}(C_{G}(y)^{0})$ is an upper bound for $d_{u}(G)$.

Proof.

Note that by Lemma 2.3(iii), $d_{q}(C_{G}(y)^{0})=d_{q}(C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0})$. Since $h=2\ell-2$, note that if $\epsilon_{v}=1$ then $\epsilon_{\alpha}=\epsilon_{\beta}$ and $\epsilon_{u}=\epsilon_{q}$, whereas if $\epsilon_{v}=0$ then $\epsilon_{u}=0$, $\epsilon_{q}=1$ and $\epsilon_{\beta}=0$. We first determine $C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0}$. Suppose that case (a) holds so that $v$ is odd. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Furthermore, since $v$ is odd, by Lemma 5.1 it follows that $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+2-\epsilon_{\beta}$ eigenvalues equal to $1$, [math] eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\lceil\beta/2\rceil$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+1}(K)}(\overline{y})^{0}=A_{\alpha}^{\lceil\frac{\beta}{2}\rceil}A_{\alpha-1}^{\frac{v-1}{2}-\lceil\frac{\beta}{2}\rceil}D_{\lceil\frac{\alpha+1}{2}\rceil}T_{\frac{v-1}{2}}$.

Suppose that case (b) holds so that $v=2$. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Now the number $N$ of eigenvalues $\omega^{i}=(-1)^{i}$ of $\overline{y}$ with $i$ odd is equal to $\ell$, $\ell-1$, $\ell+2$ or $\ell+1$ according respectively as $\ell\mod 4$ is 0, 1, 2 or 3. Since $N$ is divisible by 4, it follows from Lemma 5.1 that $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also

[TABLE]

Suppose that case (c) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (c), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\beta/2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}}A_{\alpha-1}^{\frac{v}{2}-1-\frac{\beta}{2}}D_{\frac{\alpha+1}{2}}D_{\frac{\alpha+1}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (d) or (e) holds. Note that if case (d) holds then $h\equiv 2\mod 8$ and so $\beta<v-2$ as otherwise $h=(\alpha+1)v-2$ which, under the assumptions of case (d), is equal to 6 modulo 8, a contradiction. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (d) or (e), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where if case (d) holds then $i$ is any integer with $1\leq|i|\leq\beta/2-1$ or $|i|=v/2-2$ and if case $(e)$ holds then $i$ any integer with $1\leq|i|\leq\beta/2-1$ or $|i|=v/2-1$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}}A_{\alpha-1}^{\frac{v}{2}-1-\frac{\beta}{2}}D_{\frac{\alpha+1}{2}}D_{\frac{\alpha+1}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (f) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (f), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where if $\beta\neq v-4$ then $i$ is any integer with $2\leq|i|\leq\beta/2$ or $|i|=v/2-2$ and if $\beta=v-4$ then $i$ is any integer with $1\leq|i|\leq v/2-3$ or $|i|=v/2-1$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}}A_{\alpha-1}^{\frac{v}{2}-1-\frac{\beta}{2}}D_{\frac{\alpha+1}{2}}D_{\frac{\alpha+1}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (g) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (g), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $2\leq|i|\leq\beta/2$ or $|i|=v/2-1$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}}A_{\alpha-1}^{\frac{v}{2}-1-\frac{\beta}{2}}D_{\frac{\alpha+1}{2}}D_{\frac{\alpha+1}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (h) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (h), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha+3$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq v/2-2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{v}{2}-2}A_{\alpha-1}D_{\frac{\alpha+1}{2}}D_{\frac{\alpha+3}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (i) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (i), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha-1$ eigenvalues equal to $-1$, $\alpha$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $2\leq|i|\leq v/2-1$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha+1$. Therefore $C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0}=A_{\alpha}A_{\alpha-1}^{\frac{v}{2}-2}D_{\frac{\alpha+1}{2}}D_{\frac{\alpha-1}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (j) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (j), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+1$ eigenvalues equal to $1$, $\alpha+1$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega$, $\alpha+1$ eigenvalues equal to $\omega^{-1}$, $\alpha$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $2\leq|i|\leq v/2-2$, $\alpha-1$ eigenvalues equal to $\omega^{\frac{v}{2}-1}$ and $\alpha-1$ eigenvalues equal to $\omega^{-(\frac{v}{2}-1)}$. Therefore $C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0}=A_{\alpha}A_{\alpha-1}^{\frac{v}{2}-3}A_{\alpha-2}D_{\frac{\alpha+1}{2}}D_{\frac{\alpha+1}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (k) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (k), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+2$ eigenvalues equal to $1$, $\alpha$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\beta/2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}}A_{\alpha-1}^{\frac{v}{2}-1-\frac{\beta}{2}}D_{\frac{\alpha}{2}}D_{\frac{\alpha+2}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (l) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (l), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha+2$ eigenvalues equal to $1$, $\alpha$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\beta/2-1$ or $|i|=v/2-1$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore $C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}}A_{\alpha-1}^{\frac{v}{2}-1-\frac{\beta}{2}}D_{\frac{\alpha}{2}}D_{\frac{\alpha+2}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (m) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (m), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also if $\beta\neq v-4$ then $\overline{y}$ has $\alpha+2$ eigenvalues equal to $1$, $\alpha$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $2\leq|i|\leq\beta/2$ or $|i|=v/2-2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. On the other hand if $\beta=v-4$ then $\overline{y}$ has $\alpha+2$ eigenvalues equal to $1$, $\alpha$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\beta/2-1$ or $|i|=\beta/2+1$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore in both cases $C_{{\rm SO}_{h+2}(K)}(\overline{y})^{0}=A_{\alpha}^{\frac{\beta}{2}}A_{\alpha-1}^{\frac{v}{2}-1-\frac{\beta}{2}}D_{\frac{\alpha}{2}}D_{\frac{\alpha+2}{2}}T_{\frac{v}{2}-1}$.

Suppose that case (n) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (n), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also if $\beta\neq v-2$ then $\overline{y}$ has $\alpha+2$ eigenvalues equal to $1$, $\alpha$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq\beta/2-1$ or $|i|=v/2-2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. On the other hand if $\beta=v-2$ then $\overline{y}$ has $\alpha+2$ eigenvalues equal to $1$, $\alpha+2$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq v/2-2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore

[TABLE]

Suppose that case (o) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (o), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also if $\beta\neq v-2$ then $\overline{y}$ has $\alpha+2$ eigenvalues equal to $1$, $\alpha$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $2\leq|i|\leq\beta/2$ or $|i|=v/2-1$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. On the other hand if $\beta=v-2$ then $\overline{y}$ has $\alpha+2$ eigenvalues equal to $1$, $\alpha+2$ eigenvalues equal to $-1$, $\alpha+1$ eigenvalues equal to $\omega^{i}$ where $i$ is any integer with $1\leq|i|\leq v/2-2$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore

[TABLE]

Suppose that case (p) holds. As any eigenvalue of $\overline{y}$ can be paired with its inverse, $\overline{y}$ is an element of ${\rm SO}_{h+2}(K)$ of order $v$. Let $N$ be the number of eigenvalues $\omega^{i}$ of $\overline{y}$ with $i$ odd. Then

[TABLE]

From the assumptions of case (p), it follows that $N$ is divisible by 4. Hence by Lemma 5.1, $\overline{y}$ has a preimage $y$ in $G$ of order $v$. Also $\overline{y}$ has $\alpha$ eigenvalues equal to $1$, $\alpha+2$ eigenvalues equal to $-1$, and every other eigenvalue of $\overline{y}$ occurs with multiplicity $\alpha$. Therefore

[TABLE]

By Proposition 2.2(i), $d_{u}(G)\leq d_{q}(C_{G}(y)^{0})$ and so $d_{q}(C_{G}(y)^{0})$ is an upper bound for $d_{u}(G)$. It remains to calculate $d_{q}(C_{{\rm SO}_{h+2}}(\overline{y})^{0})=d_{q}(C_{G}(y)^{0})$.

It follows from Lemma 2.4 that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

if $\alpha$ is odd then

[TABLE]

[TABLE]

and if $\alpha$ is even then

[TABLE]

Assume first that case (a) holds so that $\epsilon_{v}=1$. Recall that $\epsilon_{u}=\epsilon_{q}$ and $\epsilon_{\alpha}=\epsilon_{\beta}$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (b) holds so that $v=2$, $\epsilon_{u}=\epsilon_{v}=0$ and $\epsilon_{q}=1$. Recall that $\alpha=\ell-1$ and $\beta=0$. Note also that $e=0$ if and only if $\delta=0$. We have

[TABLE]

If $\alpha\equiv 0\mod 2$ then $z$ is even if $\delta=0$, and by Proposition 2.2(iii)

[TABLE]

If $\alpha\equiv 1\mod 4$ then by Proposition 2.2(iii)

[TABLE]

If $\alpha\equiv 3\mod 4$ then by Proposition 2.2(iii)

[TABLE]

Assume that case (c), (d), (e), (f) or (g) holds. Recall that $\epsilon_{q}=\epsilon_{\alpha}=1$ and $\epsilon_{u}=\epsilon_{v}=\epsilon_{\beta}=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (h) holds. Recall that $\epsilon_{q}=\epsilon_{\alpha}=1$, $\epsilon_{u}=\epsilon_{v}=\epsilon_{\beta}=0$ and $\beta=v-2$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (i) holds. Recall that $\epsilon_{q}=\epsilon_{\alpha}=1$, $\epsilon_{u}=\epsilon_{v}=\epsilon_{\beta}=0$ and $\beta=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (j) holds. Recall that $\epsilon_{q}=\epsilon_{\alpha}=1$, $\epsilon_{u}=\epsilon_{v}=\epsilon_{\beta}=0$ and $\beta=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (k), (l), (m), (n) or (o) holds, and that $\beta\neq v-2$ if in cases (n) or (o). Recall that $\epsilon_{q}=1$ and $\epsilon_{u}=\epsilon_{v}=\epsilon_{\alpha}=\epsilon_{\beta}=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume that case (n) or (o) holds, and $\beta=v-2$. Recall that $\epsilon_{q}=1$ and $\epsilon_{u}=\epsilon_{v}=\epsilon_{\alpha}=\epsilon_{\beta}=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

Assume finally that case (p) holds. Recall that $\epsilon_{q}=1$, $\epsilon_{u}=\epsilon_{v}=\epsilon_{\alpha}=\epsilon_{\beta}=0$ and $\beta=0$. We have

[TABLE]

By Proposition 2.2(iii),

[TABLE]

∎

7 Proofs of Theorems 4 and 5

Recall that in §4 we proved Theorem 4 for $G$ of exceptional type, see Propositions 4.2 and 4.3. In this section given a positive integer $u$ and a simple algebraic group $G$ of classical type over an algebraically closed field $K$ of characteristic $p$, we determine $d_{u}(G)$. We begin with the case where $G$ is of simply connected type and complete the proof of Theorem 4. We will then consider $G$ of neither simply connected type nor adjoint type and prove Theorem 5.

Proof of Theorem 4. As noticed at the beginning of this section, following Propositions 4.2 and 4.3 we can assume that $G$ is simply connected of classical type. By [7, Theorem 1], where Lawther determines $d_{u}(G_{a.})$ and shows that $d_{u}(G)\geq d_{u}(G_{a.})$, and Lemmas 6.1, 6.2, 6.3 and 6.4, where the upper bounds for $d_{u}(G)$ are determined accordingly respectively as $G$ is of type $A_{\ell}$, $C_{\ell}$, $B_{\ell}$, or $D_{\ell}$, we get that $d_{u}(G)=d_{u}(G_{a.})$ except possibly if one of the cases in Table 8 (see §11) holds.

It remains to show that in the cases appearing in Table 8 the upper bounds given for $d_{u}(G)-d_{u}(G_{a.})$ are in fact the precise values for $d_{u}(G)-d_{u}(G_{a.})$. As usual, we let $h$ be the Coxeter number of $G$ and write: $h=zu+e=\alpha v+\beta$ and $\alpha=zq+\delta$ where $z$, $e$, $\alpha$, $\beta$, $\delta$ are nonnegative integers such that $e<u$, $\beta<v$, and $\delta<q$. Noting that in the cases appearing in Table 8 $u$ is even, we set $s=v/2$.

Suppose first that $G$ is of type $A_{\ell}$. By Table 8, we have to consider the case where $p\neq 2$, $u$ is even, $h=zu$ and $z$ is odd. Since $e=0$, we have $\beta=\delta=0$. Also since $p\neq 2$ and $z$ is odd, $q$ and $\alpha$ are also odd.

Since $0\leq d_{u}(G)-d_{u}(G_{a.})\leq 2$, by Lemma 2.6 it suffices to show that $d_{u}(G)-d_{u}(G_{a.})>0$. Under the assumptions on $h$, $u$ and $z$, adapting the proof of of [7, Proposition 3.3], we obtain that $d_{u}(G_{a.})$ is attained for an element of $G_{a.}$ whose semisimple part $y$ of order $v$ has centralizer satisfying $C_{G_{a.}}(y)^{0}=A_{\alpha-1}^{v}T_{v-1}$. Moreover $d_{u}(G_{a.})=d_{q}(A_{\alpha-1}^{v}T_{v-1})$.

By the assumptions on $h$, $u$ and $z$ it follows that $y$ is not the image of an element of ${\rm SL}_{h}(K)$ of order dividing $v$ under the canonical map ${\rm SL}_{h}(K)\rightarrow{\rm PSL}_{h}(K)$. Hence $d_{u}(G)-d_{u}(G_{a.})>0$ and $d_{u}(G)=d_{u}(G_{a.})+2$ as required.

Suppose now that $G$ is of type $B_{\ell}$. By Table 8, we have to consider the case where $p\neq 2$, $u$ is even, $h=zu$ and $z$ is odd unless $u\equiv 2\mod 4$ and $z\equiv 2\mod 4$; moreover if $z$ is odd then either $u\equiv 2\mod 4$ and $z\equiv u/2\mod 4$, or $u\equiv 4\mod 8$. Since $e=0$, we have $\beta=\delta=0$ and $\alpha>0$. Also since $p\neq 2$, $q$ is also odd.

Write $s=v/2$ and let $a$ and $b$ be nonnegative integers such that $\ell-1=as+b$ with $0\leq b\leq s-1$. Under the assumptions on $h$, $u$ and $z$ we have $a=\alpha-1$ and $b=s-1$.

Since $0\leq d_{u}(G)-d_{u}(G_{a.})\leq 2$, by Lemma 2.6 it suffices to show that $d_{u}(G)-d_{u}(G_{a.})>0$. Under the assumptions on $h$, $u$ and $z$, adapting the proof of [7, Proposition 3.6], we obtain that $d_{u}(G_{a.})$ is attained for an element of $G_{a.}$ whose semisimple part $y$ of order $v$ has centralizer satisfying

[TABLE]

Moreover

[TABLE]

By the assumptions on $h$, $u$ and $z$ it follows from Lemma 5.1 that $y$ is not the image of an element of ${\rm Spin}_{h+1}(K)$ of order dividing $v$ under the canonical map ${\rm Spin}_{h+1}(K)\rightarrow{\rm SO}_{h+1}(K)$. Hence $d_{u}(G)-d_{u}(G_{a.})>0$ and $d_{u}(G)=d_{u}(G_{a.})+2$ as required.

Suppose now that $G$ is of type $C_{\ell}$. By Table 8, we have to consider the case where $p\neq 2$ and $u$ is even. Let $y\in G_{[v]}$ be of order $v$ and such that $Z=C_{G}(y)^{0}$ is of minimal dimension. We first show that $d_{q}(Z)-d_{u}(G_{a.})=2\lceil z/2\rceil$.

There are essentially three possibilities for the structure of $Z=C_{G}(y)^{0}$. Indeed either $y$ has no eigenvalues equal to $1$ or $-1$ in which case $Z$ has no $C$ factors, or $y$ has some eigenvalues equal to $1$ or $-1$ but not both in which case $Z$ has one $C$ factor, or $y$ has some eigenvalues equal to $1$ and $-1$ in which case $Z$ has two $C$ factors. Using Corollary 2.5 and arguing in a similar way as in the proof of [7, Proposition 3.4] we can assume that either $Z=A_{a}^{b}A_{a-1}^{s-1-b}T_{s-1}$ where $a,b$ are nonnegative integers such that $\ell=a(s-1)+b$ and $0\leq b<s-1$, or $Z=A_{a}^{\lfloor\frac{b}{2}\rfloor}A_{a-1}^{s-1-\lfloor\frac{b}{2}\rfloor}C_{\lceil\frac{a}{2}\rceil}T_{s-1}$ where $a,b$ are nonnegative integers such that $2\ell=a(2s-1)+b$ and $0\leq b<2s-1$, or $Z=A_{a}^{b-\epsilon_{a}(1-\sigma_{b})}A_{a-1}^{s-1-b+\epsilon_{a}(1-\sigma_{b})}C_{\lceil\frac{a}{2}\rceil}C_{\lceil\frac{a}{2}\rceil-\epsilon_{a}\sigma_{b}}T_{s-1}$ where $a,b$ are nonnegative integers such that $\ell=as+b$ and $0\leq b<s$.

Suppose first that $Z=A_{a}^{b}A_{a-1}^{s-1-b}T_{s-1}$ where $a,b$ are nonnegative integers such that $\ell=a(s-1)+b$ and $0\leq b<s-1$. Note that $v>2$ and so $s>1$. Write $a=cq+d$ where $c,d$ are nonnegative integers such that $0\leq d<q$. It follows from Lemma 2.4 that

[TABLE]

An easy check yields $h=zu+e=c(u-2q)+f$ where $f=d(v-2)+2b$ satisfies $0\leq f\leq u-2q-2$. In particular $c\geq z$. Write $c=z+j$ where $j\geq 0$. [7, Theorem 1] which gives the value of $d_{u}(G_{a.})$ now yields

[TABLE]

We claim that $z\geq j-1$. Suppose not. Then $c\geq 2z+2$ and so we get

[TABLE]

Since $v>2$, we have $u\geq 4q$ and so we get $0\geq 1$, a contradiction. Therefore $z\geq j-1$ as claimed. It follows that

[TABLE]

Suppose now that $Z=A_{a}^{\lfloor\frac{b}{2}\rfloor}A_{a-1}^{s-1-\lfloor\frac{b}{2}\rfloor}C_{\lceil\frac{a}{2}\rceil}T_{s-1}$ where $a,b$ are nonnegative integers such that $2\ell=a(2s-1)+b$ and $0\leq b<2s-1$. Write $a=cq+d$ where $c,d$ are nonnegative integers such that $0\leq d<q$. It follows from Lemma 2.4 that

[TABLE]

An easy check yields $h=zu+e=c(u-q)+f$ where $f=d(v-1)+b$ satisfies $0\leq f\leq u-q-1$. In particular $c\geq z$. Write $c=z+j$ where $j\geq 0$. [7, Theorem 1] which gives the value of $d_{u}(G_{a.})$ now yields

[TABLE]

We claim that $z\geq j-1$. Suppose not. Then $c\geq 2z+2$ and so we get

[TABLE]

We have $u\geq 2q$ and so we get $0\geq 1$, a contradiction. Therefore $z\geq j-1$ as claimed. It follows that

[TABLE]

Suppose finally that $Z=A_{a}^{b-\epsilon_{a}(1-\sigma_{b})}A_{a-1}^{s-1-b+\epsilon_{a}(1-\sigma_{b})}C_{\lceil\frac{a}{2}\rceil}C_{\lceil\frac{a}{2}\rceil-\epsilon_{a}\sigma_{b}}T_{s-1}$ where $a,b$ are nonnegative integers such that $\ell=as+b$ and $0\leq b<s$. Writing $a=cq+d$ where $c,d$ are nonnegative integers such that $0\leq d<q$ one can show that $c=z$ and Lemma 2.4 yields $d_{q}(Z)-d_{u}(G_{a.})=2\lceil z/2\rceil$ (see the proof of [7, Proposition 3.4]) for more details.

We have in fact showed that if $G$ is simply connected of type $C_{\ell}$, $p\neq 2$, $v$ is even and $y$ is a semisimple element of $G$ of order $v$ having centralizer $Z$ of minimal dimension then $d_{q}(Z)=d_{qv}(G_{a.})+2\lceil z/2\rceil$ where $z=\lfloor h/(qv)\rfloor$. Since

[TABLE]

and for any divisor $r$ of $v$ we have $\lfloor rh/(qv)\rfloor\geq\lfloor h/qv\rfloor$ and, by [7, Lemma 1.3], $d_{qv/r}(G_{a.})\geq d_{qv}(G_{a.})$, we deduce that $d_{u}(G)=d_{q}(Z)=d_{u}(G_{a.})+2\lceil z/2\rceil$.

Suppose finally that $G$ is of type $D_{\ell}$. By Table 8, we have to consider the case where $p\neq 2$, $z$ is odd, and $u\equiv 2\mod 4$ or $u\equiv 4\mod 8$. Moreover if $u\equiv 4\mod 8$ then $h=zu$. Finally if $u\equiv 2\mod 4$ then $h=zu$ and $z\equiv u/2\mod 4$, or $e=u-2\neq 0$ and $z\equiv 1\mod 4$. Since $v$ is even, $\beta$ is also even. Also since $p\neq 2$, $q$ is odd. Finally under the assumptions on $h$, $u$ and $z$, $\alpha$ is odd.

Write $s=v/2$ and let $a$ and $b$ be nonnegative integers such that $\ell-1=as+b$ with $0\leq b\leq s-1$. We have $a=\alpha$ is odd and $b=\beta/2$.

Suppose first that $u=2$ so that $q=1$ and $v=2$. Corollary 2.5 and Lemma 5.1 yield that an element $y$ of $G$ of order $u$ with centralizer of minimal dimension has centralizer with connected component $D_{(z+3)/2}D_{(z-1)/2}$ and $d_{u}(G)=d_{u}(G_{a.})+4$.

Suppose now that $u>2$. Then $0\leq d_{u}(G)-d_{u}(G_{a.})\leq 2$, and so by Lemma 2.6 it suffices to show that $d_{u}(G_{s.c.})-d_{u}(G_{a.})>0$. Under the assumptions on $h$, $u$ and $z$, adapting the proof of [7, Proposition 3.5], we obtain that $d_{u}(G_{a.})$ is attained for an element of $G_{a.}$ whose semisimple part $y$ of order $v$ has centralizer satisfying

[TABLE]

Moreover

[TABLE]

By the assumptions on $h$, $u$ and $z$, it follows from Lemma 5.1 that $y$ is not the image of an element of ${\rm Spin}_{h+2}(K)$ of order dividing $v$ under the canonical map ${\rm Spin}_{h+2}(K)\rightarrow{\rm SO}_{h+2}(K)$. Hence $d_{u}(G)-d_{u}(G_{a.})>0$ and $d_{u}(G_{s.c.})=d_{u}(G_{a.})+2$ as required. $\square$

Given a positive integer $u$, we can now determine $d_{u}(G)$ for $G$ a simple algebraic group of neither simply connected nor adjoint type and prove Theorem 5.

Proof of Theorem 5. Note that since $G$ is neither simply connected nor adjoint, $G$ is of classical type. We first consider part (i). If $d_{u}(G_{s.c.})=d_{u}(G_{a.})$ then Proposition 3 yields $d_{u}(G)=d_{u}(G_{a.})$.

We now consider part (ii) and assume that $d_{u}(G_{s.c.})\neq d_{u}(G_{a.})$. In particular, by Theorem 4, $p\neq 2$ and $u$ is even (and so $v$ is also even). By Proposition 3 we have $d_{u}(G_{a.})\leq d_{u}(G)\leq d_{u}(G_{s.c.})$.

(a) Suppose first that $G={\rm SL}_{\ell+1}(K)/C$ where $C\leq Z({\rm SL}_{\ell+1}(K))$ is a central subgroup of $G_{s.c.}$. Note that $C$ is finite and cyclic and write $C=\langle c\rangle$. Since $d_{u}(G_{s.c.})\neq d_{u}(G_{a.})$ Theorem 4 yields $h=zu=zqv$ where $z$ is odd.

Assume $(u,|C|)=1$. Let $gC\in G$ be an element of order dividing $u$. Then $g^{u}\in C$ and say the eigenvalue of $g^{u}$ is $c^{l}$ for some $0\leq l\leq|C|-1$. Since $(u,|C|)=1$, there is a positive integer $j$ such that $ju\equiv 1\mod|C|$. Let $k$ be a positive integer such that $k\equiv-jl\mod|C|$ and set $g^{\prime}=g\cdot{\rm diag}(c^{k},\dots,c^{k})$. Then $g^{\prime}$ is an element of $G_{s.c.}$ of order dividing $u$, $g^{\prime}C=gC$ and by Lemma 2.3 $\dim C_{G_{s.c.}}(g^{\prime})=\dim C_{G}(g^{\prime}C)$. It follows that $d_{u}(G)=d_{u}(G_{s.c.})$.

Assume now that $(u,|C|)>1$. We claim that $d_{u}(G)=d_{u}(G_{s.c.})$ if $|C|$ is odd, otherwise $d_{u}(G)=d_{u}(G_{a.})$. By Lemma 2.6, Proposition 3 and Theorem 4, $d_{u}(G)\equiv\ell\mod 2$, $d_{u}(G_{a.})\leq d_{u}(G)\leq d_{u}(G_{s.c.})$ and $d_{u}(G_{s.c})-d_{u}(G_{a.})=2$. Hence either $d_{u}(G)=d_{u}(G_{a.})$ or $d_{u}(G)=d_{u}(G_{s.c.})$.

Suppose $d_{u}(G)=d_{u}(G_{a.})$. Let $gC$ be an element of $G$ of order dividing $u$ such that $\dim C_{G}(gC)=d_{u}(G_{a.})$. Write $g=xy$ where $x$ and $y$ are respectively the unipotent and semisimple parts of $g$. As $g^{u}\in C$ and $(p,|C|)=1$, the order of $x$ divides $q$ and the order of $y$ divides $v|C|$. By Lemma 2.3 $d_{u}(G_{a.})=\dim C_{G_{s.c.}}(g)$ and so $y^{v}\neq 1$, as otherwise $g$ has order dividing $u$ and $d_{u}(G_{a})=d_{u}(G_{s.c.})$, a contradiction.

Furthermore, $y$ must have $v$ distinct eigenvalues, each repeated $zq$ times. These $v$ distinct eigenvalues must be $\omega^{1+jk},\omega^{-(1+jk)}$ where $0\leq j\leq(v-2)/2$, $k>1$ is a divisor of $|C|$ and $\omega\in K$ is a primitive $kv$-th root of unity. Considering $1\neq y^{v}\in C$ we deduce that $\omega^{v}=\omega^{-v}$ and so $\omega^{v}$ has order $2$. Hence $2$ divides $|C|$. It follows that if $|C|$ is odd then $d_{u}(G)=d_{u}(G_{s.c.})$.

Suppose that $|C|$ is even. Consider $g$ a semisimple element of $G_{s.c.}$ having $v$ distinct eigenvalues, each repeated $zq$ times, equal to $\omega^{1+2j},\omega^{-(1+2j)}$ where $0\leq j\leq(v-2)/2$, $\omega\in K$ is a primitive $2v$-th root of unity. Then $gC$ is an element of $G$ of order $v$ and $C_{G}(gC)^{0}=A_{\alpha-1}^{v}T_{v-1}$. It follows that $d_{q}(C_{G}(gC)^{0})=d_{u}(G_{a.})$ and so $d_{u}(G)=d_{u}(G_{a.})$.

(b) Suppose $G={\rm SO}_{2\ell}(K)$. Using the notation of Lemma 6.4 and Theorem 4, we let $\omega\in K$ be a primitive $v$-th root of $1$ and $y\in G$ be a semisimple element of order $v$ such that

[TABLE]

Then

[TABLE]

An easy check yields that $d_{q}(C_{G}(y))=d_{u}(G_{a.})$ and so $d_{u}(G)=d_{u}(G_{a.})$.

(c) Finally suppose that $G={\rm HSpin}_{2\ell}(K)$ where $p\neq 2$ and $\ell$ is even. Write $H=G_{s.c.}$ and let $C=Z(H)=\langle c_{1},c_{2}\rangle$, $C_{1}=\langle c_{1}\rangle$, $C_{2}=\langle c_{2}\rangle$ where $c_{1},c_{2}$ are of order 2 and $H/C=G_{a.}$, $H/C_{1}=G$ and $H/C_{2}={\rm SO}_{2\ell}(K)$.

Suppose first that $u=2$. Let $hC_{1}$ be an element of $H/C_{1}$ of order 2 satisfying

[TABLE]

By Lemma 2.3 we have

[TABLE]

Since $hC_{1}$ has order 2, either $h^{2}=1$ or $h^{2}=c_{1}$. Also note that $hC$ has order dividing 2. If $hC$ is trivial then $h\in C$ and so by (7.1) $\dim C_{H/C_{1}}(hC_{1})=\dim H/C_{1}$, a contradiction. Hence $hC$ has order 2. We claim that $h^{2}\neq c_{1}$. Suppose not. Then $hC$ is an element of $H/C$ of order 2 and $hC_{2}$ is an element of $H/C_{2}$ of order 4. It follows that $hC_{2}$ has $\ell$ eigenvalues equal to $\omega$ and $\ell$ eigenvalues equal to $\omega^{-1}$ where $\omega\in K$ is a primitive fourth root of unity. Hence $C_{H/C_{2}}(hC_{2})=A_{\ell}$. Using (7.1), it follows that $d_{2}(G)>d_{2}(G_{s.c.})$, contradicting $d_{2}(G)\leq d_{2}(G_{s.c.})$. Hence $h^{2}=1$ and so by (7.1) $d_{2}(G)\geq d_{2}(G_{s.c.})$. Therefore $d_{2}(G)=d_{2}(G_{s.c.})$.

Suppose now that $u>2.$ Since by Theorem 4 $d_{u}(G_{s.c.})-d_{u}(G_{a.})=2$, in order to show that $d_{u}(G)=d_{u}(G_{s.c.})$ it is enough by Lemma 2.6 to prove that $d_{u}(G)\neq d_{u}(G_{a.})$. Suppose for a contradiction that $d_{u}(G)=d_{u}(G_{a.})$. Let $hC_{1}$ be an element of $G=H/C_{1}$ of order $u$ such that

[TABLE]

By Lemma 2.3 we have

[TABLE]

Since $hC_{1}$ has order $u$, $h^{u}\in C_{1}$ and so $h^{u}=1$ or $h^{u}=c_{1}$. Also $hC$ has order dividing $u$ whereas $hC_{2}$ has order dividing $u$ or $2u$. Suppose $hC_{2}$ has order dividing $u$. Then $h^{u}\in C_{1}\cap C_{2}=1$ and so $h$ is an element of $G_{s.c.}$ of oder dividing $u$. Hence by (7.2), $d_{u}(G_{s.c.})=d_{u}(G_{a.})$, a contradiction.

Suppose $hC_{2}$ has order dividing $2u$ but $(hC_{2})^{u}\neq C_{2}$. Then $h^{u}=c_{1}$ and $(hC_{2})^{u}=c_{1}C_{2}$. Let $h=xy$ be the Jordan decomposition of $h$ where $x$ is unipotent and $y$ is semisimple. Note that $(yC_{2})^{v}=c_{1}C_{2}$. Without loss of generality, $yC_{2}$ is diagonal and $(yC_{2})_{i,i}(yC_{2})_{i+1,i+1}=1$ for every odd $i$ with $1\leq i\leq 2\ell-1$. Note that no eigenvalue of $yC_{2}$ is equal to $1$ or $-1$. Let $d$ be a semisimple element of $H$ such that $dC_{2}$ is diagonal, $(dC_{2})_{i,i}(dC_{2})_{i+1,i+1}=1$ for every odd $i$ with $1\leq i\leq 2\ell-1$ and $dC_{2}$ has $\ell$ eigenvalues equal to $\omega$ and $\ell$ eigenvalues equal to $\omega^{-1}$ where $\omega\in K$ is a primitive $2v$-th root of unity. Then $(dC_{2})^{v}=c_{1}C_{2}$ and $(dyC_{2})^{v}=C_{2}$. Hence $dyC_{2}$ is an element of $H/C_{2}$ of order dividing $v$ and $C_{H/C_{2}}(dyC_{2})\cong C_{H/C_{2}}(yC_{2})$. Since $\ell$ is even, it follows from Lemma 5.1 that $d$ is a semisimple element of $H$ of order dividing $2v$ and so $dy$ is a semisimple element of $H$ of order dividing $v$. Now by Lemma 2.3

[TABLE]

and so we obtain

[TABLE]

Hence $d_{u}({G_{s.c.}})=d_{u}(G_{a.})$, a contradiction. $\square$

8 Proof of Proposition 7

Given a simple algebraic group $G$ defined over an algebraically closed field $K$ of characteristic $p$, we now show that $d_{u}(G):\mathbb{N}\rightarrow\mathbb{N}$ is a decreasing function of $u$. We need to show that for every positive integer $u$, we have $d_{u}(G)-d_{u+1}(G)\geq 0$. Now by Proposition 3, we have $d_{u}(G_{a.})\leq d_{u}(G)\leq d_{u}(G_{s.c.})$ for every positive integer $u$. Hence the conclusion of the proposition will follow at once after we show that for every positive integer $u$, we have

[TABLE]

If $G$ is of exceptional type then the result follows at once from [7, Theorem 1] and Theorem 4. We therefore assume that $G$ is of classical type.

Note that if $u=1$ then $d_{u}(G_{a.})=d_{1}(G_{a.})=\dim G\geq d_{u+1}(G_{s.c.})$ and so (8.1) holds.

Suppose now that $u\geq h$. Since $u\geq h$ we have $d_{u}(G_{a.})=\ell$. Also as $u+1>h$, Theorem 4 yields $d_{u+1}(G_{s.c.})=d_{u+1}(G_{a})=\ell$. Therefore $d_{u}(G_{a.})=d_{u+1}(G_{s.c.})=\ell$ and so (8.1) holds.

We can therefore assume that $2\leq u\leq h-1$. In particular $z\geq 1$. Write $h=z^{\prime}(u+1)+e^{\prime}$ where $z^{\prime}$, $e^{\prime}$ are nonnegative integers such that $0\leq e^{\prime}\leq u.$ An easy check gives $1\leq z^{\prime}\leq z.$ Write $z^{\prime}=z-j$ where $0\leq j\leq z-1$. Then $e^{\prime}=e-z+j(u+1)$. Note that if $j=1$ then $z\geq 2$. Also $z-e\leq j(u+1)\leq u+z-e$. We claim that either $z>2j-1$ or $j=1$. Suppose otherwise. Then as $z\geq 1$ we get $j\geq 2$ and $z\leq 2j-1$. Using the latter inequality and the fact that $j(u+1)\leq u+z-e$, we get $j\leq 1-e/(u-1)$ and so $j\leq 1$, a contradiction establishing the claim.

Suppose that $G$ is of type $A$. By Theorem 4 we have $d_{u+1}(G_{s.c.})\leq d_{u+1}(G_{a.})+2$ and so

[TABLE]

Let $\mathbf{D}=d_{u}(G_{a.})-d_{u+1}(G_{a})$. Note that Lemma 2.6 yields $d_{u}(G_{a.})\equiv d_{u+1}(G_{a.})\mod 2$ and so $\mathbf{D}\equiv 0\mod 2$. Using [7, Theorem 1], we have

[TABLE]

Recalling that $z\geq 2$ if $j=1$, we note that if $j=1$ then $\mathbf{D}>0$. Also if $z>2j-1$ then $\mathbf{D}>0$.

Hence $\mathbf{D}>0$ in all cases. As $\mathbf{D}\equiv 0\mod 2$ we deduce that $\mathbf{D}\geq 2$ and so (8.2) yields $d_{u}(G_{a.})-d_{u+1}(G_{s.c.})\geq 0$. Hence (8.1) holds.

Suppose that $G$ is of type $C$. By Theorem 4 we have $d_{u+1}(G_{s.c.})\leq d_{u+1}(G_{a.})+2\epsilon_{u}\lceil(z-j)/2\rceil$ and so

[TABLE]

Let $\mathbf{D}=d_{u}(G_{a.})-d_{u+1}(G_{a})$. Note that Lemma 2.6 yields $d_{u}(G_{a.})\equiv d_{u+1}(G_{a.})\mod 2$ and so $\mathbf{D}\equiv 0\mod 2$. Using [7, Theorem 1], we have

[TABLE]

Recalling that $z\geq 2$ if $j=1$, we note that if $j=1$ then $\mathbf{D}\geq 0$ and moreover $\mathbf{D}\geq(z^{2}+\epsilon_{z})/2\geq z+\epsilon_{z}=2\lceil z/2\rceil$ for $u$ odd. Also if $z>2j-1$ then $\mathbf{D}\geq 0$ and moreover $\mathbf{D}\geq 2\lceil z/2\rceil$ for $u$ odd.

In all cases we therefore have $\mathbf{D}\geq 0$ and moreover $\mathbf{D}\geq 2\lceil z/2\rceil$ for $u$ odd. Now (8.3) yields $d_{u}(G_{a.})-d_{u+1}(G_{s.c.})\geq 0$. Hence (8.1) holds.

Suppose that $G$ is of type $B$. By Theorem 4 we have $d_{u+1}(G_{s.c.})\leq d_{u+1}(G_{a.})+2\epsilon_{u}$ and so

[TABLE]

Let $\mathbf{D}=d_{u}(G_{a.})-d_{u+1}(G_{a})$. Note that Lemma 2.6 yields $d_{u}(G_{a.})\equiv d_{u+1}(G_{a.})\mod 2$ and so $\mathbf{D}\equiv 0\mod 2$. Using [7, Theorem 1], we have

[TABLE]

Arguing as in the case where $G$ is of type $C$, we deduce that $\mathbf{D}\geq 0$ and moreover $\mathbf{D}\geq 2$ for $u$ odd. Now (8.4) yields $d_{u}(G_{a.})-d_{u+1}(G_{s.c.})\geq 0$. Hence (8.1) holds.

Suppose that $G$ is of type $D$. Since $u\geq 2$, by Theorem 4 we have $d_{u+1}(G_{s.c.})\leq d_{u+1}(G_{a.})+2\epsilon_{u}$ and so

[TABLE]

Let $\mathbf{D}=d_{u}(G_{a.})-d_{u+1}(G_{a})$. Note that Lemma 2.6 yields $d_{u}(G_{a.})\equiv d_{u+1}(G_{a.})\mod 2$ and so $\mathbf{D}\equiv 0\mod 2$. Using [7, Theorem 1], we have

[TABLE]

Arguing as in the case where $G$ is of type $B$, we deduce that $\mathbf{D}\geq 0$ and moreover $\mathbf{D}\geq 2$ for $u$ odd. We deduce that $d_{u}(G_{a.})-d_{u+1}(G_{s.c.})\geq 0$. Hence (8.1) holds. $\square$

9 Proof of Theorem 9

In this section we prove Theorem 9. We proceed in two steps, first in the special case where $G$ is of simply connected type (see Proposition 9.1), and then we consider other types (see Proposition 9.2).

Proposition 9.1**.**

Let $G$ be a simple simply connected algebraic group over an algebraic closed field $K$ of prime characteristic $p$. The classification of the hyperbolic triples $(a,b,c)$ of integers for $G$ is as given in Theorem 9.

Proof.

As usual we let $h$ be the Coxeter number of $G$ and for a positive integer $u$, we let $z$ and $e$ be the nonnegative integers such that $h=zu+e$ and $0\leq e\leq u-1$. Also $d_{u}(G)$ denotes the minimal dimension of the centralizer of an element of $G$ of order dividing $u$. By Proposition 2 $d_{u}(G)$ is the codimension of the subvariety $G_{[u]}$ of $G$ consisting of elements of order dividing $u$. Writing $j_{u}(G)=\dim G_{[u]}$, we have $d_{u}(G)=\dim G-j_{u}(G)$.

Let $(a,b,c)$ be a hyperbolic triple of integers. Let $\mathbf{S}_{(a,b,c)}=d_{a}(G)+d_{b}(G)+d_{c}(G)$ and $\mathbf{D}_{(a,b,c)}=\mathbf{S}_{(a,b,c)}-\dim G$. Saying that $(a,b,c)$ is reducible (respectively, rigid, nonrigid) for $G$ amounts to saying that $\mathbf{D}_{(a,b,c)}$ is greater than (respectively equal to, less than) 0.

Recall the partial order we put on the set of hyperbolic triples of integers. Given two hyperbolic triples $(a,b,c)$ and $(a^{\prime},b^{\prime},c^{\prime})$ of integers, we say that $(a,b,c)\leq(a^{\prime},b^{\prime},c^{\prime})$ if and only if $a\leq a^{\prime}$, $b\leq b^{\prime}$ and $c\leq c^{\prime}$. Among all hyperbolic triples, exactly three are minimal: $(2,3,7)$, $(3,3,4)$ and $(2,4,5)$. By Proposition 7, if $(a,b,c)$ is nonrigid for $G$ then $(a^{\prime},b^{\prime},c^{\prime})$ is nonrigid for $G$ for every $(a^{\prime},b^{\prime},c^{\prime})\geq(a,b,c)$.

For a nonnegative integer $p$, we let $\theta_{p}\in\{0,1\}$ be such that $\theta_{p}=1$ if $p$ is odd or $p=0$, otherwise $\theta_{p}=0$.

Suppose first that $G$ is of exceptional type. By Theorem 4 and [7, Theorem 1], every hyperbolic triple $(a,b,c)$ of integers is nonrigid for $G$ unless $G$ is of type $G_{2}$ and $(a,b,c)\in\{(2,4,5),(2,5,5)\}$ in which case $(a,b,c)$ is rigid for $G$.

Suppose now that $G$ is of classical type. In a first step we show that if $\ell\geq 11$, $\ell\geq 10$, $\ell\geq 15$, or $\ell\geq 9$ accordingly respectively as $G$ is of type $A$, $B$, $C$ or $D$, then every hyperbolic triples of integers is nonrigid for $G$.

Suppose first that $G$ is of type $A$. Note that $\dim G=h^{2}-1$. By Theorem 4 and [7, Theorem 1], we have

[TABLE]

Let $g(e)=\frac{(h-e)(h+e)}{u}+e-1+2\theta_{p}(1-\epsilon_{u}).$ Then $g^{\prime}(e)=\frac{u-2e}{u}$, and $g^{\prime}(e)>0$ if and only if $e<u/2$. Hence $d_{u}(G)\leq F(u)$ where

[TABLE]

In particular, for any hyperbolic triple $(a,b,c)$ of integers, we have $\mathbf{S}_{(a,b,c)}\leq F(a)+F(b)+F(c)$.

Suppose $(a,b,c)=(2,3,7)$. We have $\mathbf{S}_{(2,3,7)}\leq F(2)+F(3)+F(7)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{h^{2}}{42}+2\theta_{p}+1$ is negative for $h\geq 12$, it follows that $(2,3,7)$ is nonrigid for $G$ provided $h\geq 12$, that is $\ell\geq 11$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(2,3,7)$ is nonrigid for $G$ with $\ell\geq 11$.

Suppose $(a,b,c)=(2,4,5)$. We have $\mathbf{S}_{(2,4,5)}\leq F(2)+F(4)+F(5)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{h^{2}}{20}+4\theta_{p}+\frac{3}{4}$ is negative for $h\geq 10$, it follows that $(2,4,5)$ is nonrigid for $G$ provided $h\geq 10$, that is $\ell\geq 9$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(2,4,5)$ is nonrigid for $G$ with $\ell\geq 9$.

Suppose $(a,b,c)=(3,3,4)$. We have $\mathbf{S}_{(3,3,4)}\leq F(3)+F(3)+F(4)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{h^{2}}{12}+2\theta_{p}+\frac{1}{2}$ is negative for $h\geq 6$, it follows that $(3,3,4)$ is nonrigid for $G$ provided $h\geq 6$, that is $\ell\geq 5$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(3,3,4)$ is nonrigid for $G$ with $\ell\geq 5$.

Suppose now that $G$ is of type $B$. Note that $\dim G=h(h+1)/2$. By Theorem 4 and [7, Theorem 1], we have

[TABLE]

Let $g(e)=\frac{(h-e)(h+e+1)}{2u}+\frac{e+1}{2}+2\theta_{p}(1-\epsilon_{u}).$ Then $g^{\prime}(e)=\frac{u-2e-1}{2u}$, and $g^{\prime}(e)>0$ if and only if $e<(u-1)/2$. Hence $d_{u}(G)\leq F(u)$ where

[TABLE]

In particular, for any hyperbolic triple $(a,b,c)$ of integers, we have $\mathbf{S}_{(a,b,c)}\leq F(a)+F(b)+F(c)$.

Suppose $(a,b,c)=(2,3,7)$. We have $\mathbf{S}_{(2,3,7)}\leq F(2)+F(3)+F(7)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{4h^{2}+4h-797}{336}+2\theta_{p}$ is negative for $h\geq 20$, it follows that $(2,3,7)$ is nonrigid for $G$ provided $h\geq 20$, that is $\ell\geq 10$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(2,3,7)$ is nonrigid for $G$ with $\ell\geq 10$.

Suppose $(a,b,c)=(2,4,5)$. We have $\mathbf{S}_{(2,4,5)}\leq F(2)+F(4)+F(5)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{4h^{2}+4h-359}{160}+4\theta_{p}$ is negative for $h\geq 16$, it follows that $(2,4,5)$ is nonrigid for $G$ provided $h\geq 16$, that is $\ell\geq 8$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(2,4,5)$ is nonrigid for $G$ with $\ell\geq 8$.

Suppose $(a,b,c)=(3,3,4)$. We have $\mathbf{S}_{(3,3,4)}\leq F(3)+F(3)+F(4)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{4h^{2}+4h-203}{96}+2\theta_{p}$ is negative for $h\geq 10$, it follows that $(3,3,4)$ is nonrigid for $G$ provided $h\geq 10$, that is $\ell\geq 5$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(3,3,4)$ is nonrigid for $G$ with $\ell\geq 5$.

Suppose now that $G$ is of type $C$. Note that $\dim G=h(h+1)/2$. By Theorem 4 and [7, Theorem 1], we have

[TABLE]

Let

[TABLE]

Then

[TABLE]

and if $u$ is even then $g^{\prime}(e)>0$ if and only if $e<(u-2)/2$, else $g^{\prime}(e)>0$ if and only if $e<(u-1)/2$. Hence $d_{u}(G)\leq F(u)$ where

[TABLE]

In particular, for any hyperbolic triple $(a,b,c)$ of integers, we have $\mathbf{S}_{(a,b,c)}\leq F(a)+F(b)+F(c)$.

Suppose $(a,b,c)=(2,3,7)$. We have $\mathbf{S}_{(2,3,7)}\leq F(2)+F(3)+F(7)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{h^{2}-20h-236}{84}$ is negative for $h\geq 30$, it follows that $(2,3,7)$ is nonrigid for $G$ provided $h\geq 30$, that is $\ell\geq 15$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(2,3,7)$ is nonrigid for $G$ with $\ell\geq 15$.

Suppose $(a,b,c)=(2,4,5)$. We have $\mathbf{S}_{(2,4,5)}\leq F(2)+F(4)+F(5)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{h^{2}-14h-121}{40}$ is negative for $h\geq 22$, it follows that $(2,4,5)$ is nonrigid for $G$ provided $h\geq 22$, that is $\ell\geq 11$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(2,4,5)$ is nonrigid for $G$ with $\ell\geq 11$.

Suppose $(a,b,c)=(3,3,4)$. We have $\mathbf{S}_{(3,3,4)}\leq F(3)+F(3)+F(4)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{h^{2}-2h-59}{24}$ is negative for $h\geq 10$, it follows that $(3,3,4)$ is nonrigid for $G$ provided $h\geq 10$, that is $\ell\geq 5$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(3,3,4)$ is nonrigid for $G$ with $\ell\geq 5$.

Suppose finally that $G$ is of type $D$. Note that $\dim G=(h+1)(h+2)/2$. By Theorem 4 and [7, Theorem 1], we have

[TABLE]

Let

[TABLE]

Then

[TABLE]

and if $u$ is even then $g^{\prime}(e)>0$ if and only if $e<(u-2)/2$, else $g^{\prime}(e)>0$ if and only if $e<(u-3)/2$. Hence $d_{u}(G)\leq F(u)$ where

[TABLE]

In particular, for any hyperbolic triple $(a,b,c)$ of integers, we have $\mathbf{S}_{(a,b,c)}\leq F(a)+F(b)+F(c)$.

Suppose $(a,b,c)=(2,3,7)$. We have $\mathbf{S}_{(2,3,7)}\leq F(2)+F(3)+F(7)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{h^{2}+24h-276}{84}+4\theta_{p}$ is negative for $h\geq 16$, it follows that $(2,3,7)$ is nonrigid for $G$ provided $h\geq 16$, that is $\ell\geq 9$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(2,3,7)$ is nonrigid for $G$ with $\ell\geq 9$.

Suppose $(a,b,c)=(2,4,5)$. We have $\mathbf{S}_{(2,4,5)}\leq F(2)+F(4)+F(5)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{h^{2}+18h-109}{40}+6\theta_{p}$ is negative for $h\geq 12$, it follows that $(2,4,5)$ is nonrigid for $G$ provided $h\geq 12$, that is $\ell\geq 7$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(2,4,5)$ is nonrigid for $G$ with $\ell\geq 7$.

Suppose $(a,b,c)=(3,3,4)$. We have $\mathbf{S}_{(3,3,4)}\leq F(3)+F(3)+F(4)$. Now

[TABLE]

Hence

[TABLE]

and

[TABLE]

Since $-\frac{h^{2}+6h-75}{24}+2\theta_{p}$ is negative for $h\geq 10$, it follows that $(3,3,4)$ is nonrigid for $G$ provided $h\geq 10$, that is $\ell\geq 6$. Also every hyperbolic triple $(a^{\prime},b^{\prime},c^{\prime})$ of integers with $(a^{\prime},b^{\prime},c^{\prime})\geq(3,3,4)$ is nonrigid for $G$ with $\ell\geq 6$.

Finally to classify hyperbolic triples $(a,b,c)$ of integers for $G$ with $\ell\leq 10$, $\ell\leq 9$, $\ell\leq 14$, $\ell\leq 8$ accordingly respectively as $G$ is of type $A$, $B$, $C$, $D$ one can use Theorem 4 and [7, Theorem 1] to find $d_{a}(G)$, $d_{b}(G)$ and $d_{c}(G)$ and compute $\mathbf{D}_{(a,b,c)}$. Note that if $u>h$ then $d_{u}(G)=\ell$. ∎

Proposition 9.2**.**

Let $G$ be a simple algebraic group over an algebraic closed field $K$ of prime characteristic $p$. The classification of hyperbolic triples $(a,b,c)$ of integers for $G$ is as given in Theorem 9.

Proof.

The case where $G$ is of simply connected type is treated in Proposition 9.1. We therefore suppose that $G$ is not of simply connected type. We denote by $G_{s.c.}$ (respectively, $G_{a.}$) the simple algebraic group of simply connected (respectively, adjoint) type having the same Lie type and Lie rank as $G$.

For an integer $u$, let $d_{u}(G)$ denote the minimal dimension of the centralizer of an element of $G$ of order dividing $u$. By Proposition 2 $d_{u}(G)$ is the codimension of the subvariety $G_{[u]}$ of $G$ consisting of elements of order dividing $u$. Writing $j_{u}(G)=\dim G_{[u]}$, we have $d_{u}(G)=\dim G-j_{u}(G)$.

Let $(a,b,c)$ be a hyperbolic triple of integers. Let $\mathbf{S}_{(a,b,c)}=d_{a}(G)+d_{b}(G)+d_{c}(G)$ and $\mathbf{D}_{(a,b,c)}=\mathbf{S}_{(a,b,c)}-\dim G$. Recall that saying that $(a,b,c)$ is reducible (respectively, rigid, nonrigid) for $G$ amounts to saying that $\mathbf{D}_{(a,b,c)}$ is greater than (respectively equal to, less than) 0.

By Proposition 3, given a positive integer $u$, we have $d_{u}(G)\leq d_{u}(G_{s.c.})$. It follows that every hyperbolic triple of integers which is nonrigid for $G_{s.c.}$ is nonrigid for $G$. In particular, by Proposition 9.1, we can now assume that $G$ is of classical type. By the proof of Proposition 9.1 if $G$ is such that $\ell\leq 11$, $\ell\leq 10$, $\ell\leq 15$, $\ell\leq 9$ accordingly respectively as $G$ is of type $A$, $B$, $C$, $D$ then every hyperbolic triple of integers is nonrigid for $G$. For $G$ of small rank, one can use [7, Theorem 1] and Theorems 4 and 5 to find $d_{a}(G)$, $d_{b}(G)$ and $d_{c}(G)$, compute $\mathbf{D}_{(a,b,c)}$ and classify a given hyperbolic triple $(a,b,c)$ of integers for $G$. Note that if $u>h$ then $d_{u}(G)=\ell$. ∎

10 Proof of Proposition 10

We consider Proposition 10. Let $G$ be a simple algebraic group over an algebraically closed field $K$. The main ingredient in the proof of Proposition 10 is the following result proved in [10] combined with the classification given in Theorem 9 of the reducible and the rigid hyperbolic triples of integers for simple algebraic groups if $p$ is a bad prime for $G$ or $G$ is of exceptional type. Recall that $p$ is said to be bad for $G$ if $G$ is of type $B_{\ell}$, $C_{\ell}$, $D_{\ell}$ and $p=2$, or of type $G_{2}$, $F_{4}$, $E_{6}$, $E_{7}$ and $p\in\{2,3\}$, or of type $E_{8}$ and $p\in\{2,3,5\}$. A prime $p$ is said to be good for $G$ if it is not bad for $G$. Also in the statement below and from now on, by an irreducible subgroup of a classical group $G$, we mean a subgroup acting irreducibly on the natural module for $G$.

Proposition 10.1**.**

[10, Proposition 2.1]**. Suppose that $G$ is of classical type and $p$ is a good prime for $G$. If $g_{1}$, $g_{2}$, $g_{3}$ are elements of $G$ such that $g_{1}g_{2}g_{3}=1$ and $\langle g_{1},g_{2}\rangle$ is an irreducible subgroup of $G$ then

[TABLE]

We can now prove Proposition 10.

Proof of Proposition 10. If $G$ is of classical type and $p$ is a good prime for $G$ then the result follows from Proposition 10.1. The remaining cases follow from Theorem 9 which shows that there are no reducible hyperbolic triples of integers for $G$ if $G$ is of symplectic or orthogonal type and $p=2$, or if $G$ is of exceptional type. $\hskip 142.26378pt\square$

11 Some tables

In this section we collect Tables 4-8 of the paper. Tables 4-7 appear in §6 and Table 8 appears in §7.

[FIGURE:]

[FIGURE:]