Presentation of finite subgroups of mapping class group of genus 2 surface by Dehn twists

TL;DR

This paper provides explicit presentations of all finite subgroups of the genus 2 surface's mapping class group using Humphries generators, up to conjugacy.

Contribution

It offers a complete classification and presentation of finite subgroups of the genus 2 mapping class group, a previously unlisted comprehensive result.

Findings

Presentations of all finite subgroups by Humphries generators

Classification of subgroups up to conjugacy

Explicit descriptions for genus 2 surface

Abstract

In this note we give presentations of all finite subgroups of the mapping class group of a closed surface of genus 2 by the Humphries generators up to conjugacy.

Equations

$$\displaystyle\omega_{i}\omega_{j}=\omega_{j}\omega_{i}\quad\mbox{if $|i-j|\geq 2$, $1\leq i,j\leq 2g+1$}$$

$${\omega }_i{\omega }_{i+1}{\omega }_i={\omega }_{i+1}{\omega }_i{\omega }_{i+1}(1\le i\le 2g)$$

$$({\omega }_1{\omega }_2\cdots {\omega }_{2g+1}{)}^{2g+2}=1\text{mbox}(achainrelation)$$

$${\zeta }_0={\omega }_1{\omega }_2\cdots {\omega }_{2g}{\omega }_{2g+1}{\omega }_{2g+1}{\omega }_{2g}\cdots {\omega }_2{\omega }_1$$

$$\begin{array}[]{l}\omega_{i+k}(\omega_{i}\omega_{i+1}\cdots\omega_{i+m})=(\omega_{i}\omega_{i+1}\cdots\omega_{i+m})\omega_{i+k-1},\\ \\ \omega_{i+k}^{-1}(\omega_{i}^{-1}\omega_{i+1}^{-1}\cdots\omega_{i+m}^{-1})=(\omega_{i}^{-1}\omega_{i+1}^{-1}\cdots\omega_{i+m}^{-1})\omega_{i+k-1}^{-1},\end{array}$$

$${\omega }_{i+k}({\omega }_i{\omega }_{i+1}\cdots {\omega }_{i+m})$$

$$\zeta ={\omega }_1{\omega }_2\cdots {\omega }_{2g+1},\eta ={\omega }_1{\omega }_2\cdots {\omega }_{2g}.$$

$${\omega }_1\zeta =\zeta {\zeta }^{-1}{\omega }_1\zeta =\zeta {\omega }_{2g+1}^{-1}{\omega }_{2g}^{-1}\cdots {\omega }_3^{-1}{\omega }_2^{-1}\zeta ={\zeta }^2{\omega }_{2g}^{-1}\cdots {\omega }_2^{-1}{\omega }_1^{-1},$$

$${\omega }_i{\zeta }^j={\zeta }^j{\omega }_{i-j}(i\ne j),{\omega }_i{\zeta }^i={\zeta }^{i+1}{\eta }^{-1},$$

$${\omega }_i={\zeta }^{i+1}{\eta }^{-1}{\zeta }^{-i},(i=1,2,...,2g+1).$$

$${\eta }^{2g+1}$$

$$\xi^{2g}=\underbrace{\omega_{1}\eta\omega_{1}\eta\cdots\omega_{1}\eta}_{\mbox{$2g$ times}}=\omega_{1}\omega_{2}\cdots\omega_{2g}\eta^{2g}=\eta^{2g+1}.$$

$${\zeta }_1^6=({\omega }_1{\omega }_2{\omega }_3{\omega }_4{\omega }_5{)}^6=1,$$

$${\zeta }_0^2=({\omega }_1{\omega }_2{\omega }_3{\omega }_4{\omega }_5^2{\omega }_4{\omega }_3{\omega }_2{\omega }_1{)}^2=1,$$

$${\omega }_1{\omega }_2{\omega }_3{\omega }_4{\omega }_5^2{\omega }_4{\omega }_3{\omega }_2{\omega }_1$$

$${\omega }_i{\zeta }_1^j={\zeta }_1^j{\omega }_{i-j}(i\ne j),{\omega }_i{\zeta }_1^i={\zeta }_1^{i+1}{\zeta }_4^{-1},$$

$${\zeta }_1{\omega }_5={\zeta }_0{\omega }_1^{-1}{\omega }_2^{-1}{\omega }_3^{-1}{\omega }_4^{-1},$$

$${\omega }_1^{-1}{\omega }_2^{-1}{\omega }_3^{-1}{\omega }_4^{-1}{\omega }_5^{-1}={\zeta }_1{\zeta }_0^{-1}={\zeta }_1{\zeta }_0.$$

$${\zeta }_2$$

$$ba={\omega }_1{\omega }_2{\omega }_5^{-1}{\omega }_4^{-1}{\zeta }_1^3={\omega }_1{\omega }_2{\zeta }_1^3{\omega }_2^{-1}{\omega }_1^{-1}.$$

$$G_n=⟨x,z=y^2:x^2=z^4=1,xz{x}^{-1}=z^{-1}⟩.$$

$$xy{x}^{-1}={\zeta }_1^3({\omega }_1{\omega }_2{\omega }_4{\omega }_3{\omega }_2){\zeta }_1^{-3}$$

$$⟨x,y,u:x^2=y^2=u^6=[x,y]=[y,u]=1,xu{x}^{-1}=u^{-1}y⟩.$$

$$G_r=⟨t,w:t^4=w^3=1,tw{t}^{-1}=w^{-1}⟩$$

$${\zeta }_5$$

$${\zeta }_5^2={\zeta }_0,{\zeta }_5^{-1}{\zeta }_1{\zeta }_5={\zeta }_1^{-1}{\zeta }_0.$$

$$(\zeta_{5}\zeta_{1}^{k})^{2}=\zeta_{0}\ \mbox{if $k$ is even},\ (\zeta_{5}\zeta_{1}^{k})^{2}=1\ \mbox{if $k$ is odd}.$$

$$x=\left(\begin{array}[]{cc}1&1\\ 0&1\end{array}\right),\quad y=\left(\begin{array}[]{cc}0&1\\ -1&0\end{array}\right)\quad\mbox{(the entries are in $\mathbb{Z}/(3\mathbb{Z}))$}.$$

$$u=\left(\begin{array}[]{cc}0&-1\\ -1&-1\end{array}\right),$$

$$u^2=xy{x}^{-1}{y}^2,ux{u}^{-1}=y^{-1}{x}^{-1}y,uy{u}^{-1}=x^{-1}yx.$$

$$G_m=⟨v,y:v^4=y^4=1,v^2=y^2,vy{v}^{-1}=y^{-1}⟩.$$

$${\zeta }_2{\zeta }_3^{-2}$$

$$\begin{array}[]{l}a=\omega_{1}^{-1}\zeta_{2}\omega_{1}=\omega_{2}\omega_{4}^{-1}\omega_{5}^{-1}\omega_{1}=\omega_{2}\omega_{1}\omega_{4}^{-1}\omega_{5}^{-1},\\ b=\omega_{1}^{-1}\zeta_{3}\omega_{1}=\omega_{1}\omega_{2}\omega_{3}\omega_{4}\omega_{1}=\zeta_{4}\omega_{1}=\omega_{2}\omega_{1}\omega_{2}\omega_{3}\omega_{4},\\ c=\omega_{2}\omega_{3}\omega_{5}\omega_{4}\omega_{3}=(\zeta_{1}\omega_{4}\omega_{2}^{-1}\omega_{1}^{-1})\zeta_{3}(\zeta_{1}\omega_{4}\omega_{2}^{-1}\omega_{1}^{-1})^{-1}\end{array}$$

Full text

1 Introduction

Let ${\cal MCG}_{g}$ denote the mapping class group of a closed orientable surface of genus $g$. The Dehn-Lickorish theorem states that ${\cal MCG}_{g}$ is generated by Dehn twists along finitely many simple closed curves. S. P. Humphries found $2g+1$ Dehn twists which generate ${\cal MCG}_{g}$ for $g>1$. A finite presentation in the Humphries generators was given by Wajnryb [7].

The objective of this note is to give a generator system represented by products of Humphries generators $\{\omega_{j}\}_{j=1}^{5}$ for all finite subgroups, up to topological conjugacy, of the mapping class group ${\cal MCG}_{2}$ of genus $2$. In general, it is not easy to find finite subgroups of a group when only one of its presentations is given. For ${\cal MCG}_{g}$, due to the Nielsen realization theorem (S. Kerckhoff [5]), its finite subgroup is represented by a group of holomorphic automorphisms on a closed Riemann surface of genus $g$. S. A. Broughton [2] made a complete list of all groups, up to topological conjugacy, which arise as groups of holomorphic automorphisms on some closed Riemann surfaces of genus $2$. It is natural to ask how those groups are embedded in $\mathcal{MCG}_{2}$.

Our main theorem represents generators by products of $\omega_{i}$ ($i=1,...,5$) for each group in Broughton’s list. For its statement we introduce some notation. Let a finite group $G$ act on a Riemann surface $R$ as a group of holomorphic automorphisms. If the genus of the factor surface $R/G$ is $h$ and the covering map $\pi:R\to R/G$ is branched over $n$ points $p_{1}$,…, $p_{n}$ with branching orders $m_{j}$,then $(h;m_{1},...,m_{n})$ is the type of the orbifold $R/G$. In stead of $(h;m_{1},...,m_{n})$, we often write $(h;\nu_{1}^{r_{1}},...,\nu_{p}^{r_{p}})$ if $\nu_{j}$ appears $r_{j}$ times in $(m_{1},...,m_{n})$. Let $\zeta_{0}$, $\zeta_{1}$, …, $\zeta_{4}$ be as in the following table. They have the orders indicated in the table.

[TABLE]

The list below shows the group $G_{\ast}$ corresponding to (2.$\ast$) in [2], the order $|G_{\ast}|$ and the orbifold type.

Theorem 1.1

A non-trivial finite subgroup of ${\cal MCG}_{2}$ of a closed orientable surface of genus $2$ is conjugate with one of the groups in the following list.

• (2.a)

$G_{a}=\langle x=\zeta_{0}:x^{2}=1\rangle\cong\mathbb{Z}_{2}$, $2$,  $(0;2^{6})$.

• (2.b)

$G_{b}=\langle x=\zeta_{1}^{3}:x^{2}=1\rangle\cong\mathbb{Z}_{2}$, $2$,  $(1;2^{2})$.

• (2.c)

$G_{c}=\langle x=\zeta_{1}^{2}:x^{3}=1\rangle\cong\mathbb{Z}_{3}$, $3$,  $(0;3^{4})$.

• (2.e)

$G_{e}=\langle x=\zeta_{3}^{2}:x^{4}=1\rangle\cong\mathbb{Z}_{4}$, $4$,  $(0;2^{2},4^{2})$.

• (2.f)

$G_{f}=\langle x=\zeta_{0},y=\zeta_{1}^{3}:x^{2}=y^{2}=[x,y]=1\rangle\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$, $4$,  $(0;2^{5})$.

• (2.h)

$G_{h}=\langle x=\zeta_{4}^{2}:x^{5}=1\rangle\cong\mathbb{Z}_{5}$, $5$,  $(0;5^{3})$.

• (2.i)

$G_{i}=\langle x=\zeta_{1}:x^{6}=1\rangle\cong\mathbb{Z}_{6}$, $6$,  $(0;3,6^{2})$.

• (2.k1)

$G_{k1}=\langle x=\zeta_{2}:x^{6}=1\rangle\cong\mathbb{Z}_{6}$, $6$,  $(0;2^{2},3^{2})$.

• (2.k2)

$G_{k2}=\langle x=\zeta_{1}^{3},y=(\omega_{4}\zeta_{2}\omega_{4}^{-1})^{2}:x^{2}=y^{3}=1,xyx^{-1}=y^{-1}\rangle\cong D_{3}$, $6$,  $(0;2^{2},3^{2})$.

• (2.l)

$G_{l}=\langle x=\zeta_{3}:x^{8}=1\rangle\cong\mathbb{Z}_{8}$, $8$,  $(0;2,8,8)$.

• (2.m)

$G_{m}=\langle x=\zeta_{2}^{-1}\zeta_{3}^{2}\zeta_{2},y=\zeta_{3}^{2}:x^{4}=y^{4}=1,x^{2}=y^{2},xyx^{-1}=y^{-1}\rangle\cong\tilde{D}_{2}$, $8$,  $(0;4,4,4)$.

• (2.n)

$G_{n}=\langle x,y:x^{2}=y^{4}=1,xyx^{-1}=y^{-1}\rangle\cong D_{4}$, where $x=\zeta_{1}^{3}$, $y=(\omega_{4}\omega_{2}^{-1}\omega_{1}^{-1})\zeta_{3}^{2}(\omega_{4}\omega_{2}^{-1}\omega_{1}^{-1})^{-1}$, $8$,  $(0;2^{3},4)$.

• (2.o)

$G_{o}=\langle x=\zeta_{4}:x^{10}=1\rangle\cong\mathbb{Z}_{10}$, $10$,  $(0;2,5,10)$.

• (2.p)

$G_{p}=\langle x=\zeta_{0},y=\zeta_{1}:x^{2}=y^{6}=[x,y]=1\rangle\cong\mathbb{Z}_{2}\times\mathbb{Z}_{6}$, $12$,  $(0;2;6,6)$.

• (2.r)

$G_{r}=\langle x,y:x^{4}=y^{3}=1,xyx^{-1}=y^{-1}\rangle\cong D_{4,3,-1}$,

where $x=(\omega_{3}\omega_{5}^{-1}\omega_{1}^{-1})\zeta_{3}^{6}(\omega_{3}\omega_{5}^{-1}\omega_{1}^{-1})^{-1},y=\zeta_{1}^{4}$, $12$,  $(0;3,4^{2})$.

• (2.s)

$G_{s}=\langle x=\zeta_{1}^{3},y=\omega_{4}\zeta_{2}\omega_{4}^{-1}:x^{2}=y^{6}=1,xyx^{-1}=y^{-1}\rangle\cong D_{6}$, $12$,  $(0;2^{3},3)$.

• (2.u)

$G_{u}=\langle x,y:x^{2}=y^{8}=1,xyx^{-1}=y^{3}\rangle\cong D_{2,8,3}$, where $x=\zeta_{1}^{3},y=(\omega_{4}\omega_{2}^{-1}\omega_{1}^{-1})\zeta_{3}(\omega_{4}\omega_{2}^{-1}\omega_{1}^{-1})^{-1}$, 16,  $(0;2,4,8)$.

• (2.w)

G_{w}=\left\langle x,y,z,w:\begin{array}[]{l}x^{2}=y^{2}=z^{2}=w^{3}=[y,z]=[y,w]=[z,w]=1\\ xyx^{-1}=y,xzx^{-1}=zy,xwx^{-1}=w^{-1}\end{array}\right\rangle,

where $x=(\omega_{1}\omega_{2}\omega_{1})\zeta_{1}^{3}(\omega_{1}\omega_{2}\omega_{1})^{-1}$, $y=\zeta_{0},z=\zeta_{1}^{3},w=\zeta_{1}^{4}$,

$G_{w}\cong\mathbb{Z}_{2}\ltimes(\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{3})$, $24$,  $(0;2,4,6)$.

• (2.x)

$G_{x}=\langle x=\zeta_{2}^{4},y=\zeta_{3}^{2}:x^{3}=y^{4}=1,xy^{2}=y^{2}x,(xy)^{3}=1\rangle\cong SL_{2}(3)$, $24$,  $(0;3^{2},4)$

• (2.aa)

G_{aa}=\left\langle x,y,u:\begin{array}[]{l}x^{3}=y^{4}=(xy)^{3}=1,xy^{2}=y^{2}x,u^{2}=xyx^{-1}y^{2}\\ uxu^{-1}=y^{-1}x^{-1}y,uyu^{-1}=x^{-1}yx\end{array}\right\rangle

where $x=\zeta_{2}^{4},y=\zeta_{3}^{2}$, $u=(\omega_{1}\zeta_{1}\omega_{4}\omega_{2}\omega_{1}^{-1})\zeta_{3}(\omega_{1}\zeta_{1}\omega_{4}\omega_{2}\omega_{1}^{-1})^{-1}$,

$G_{aa}\cong GL_{2}(3)$, $48$,  $(0;2,3,8)$

We thank Susumu Hirose and Shigeru Takamura for their encouragement and valuable comments.

2 Facts about mapping class groups

2.1 The mapping class group and its generators

Our basic reference for mapping class groups is Farb-Margalit’s book [3], in particular, Sections 3, 4 and 7. The mapping class group ${\cal MCG}_{g}$ is generated by the Humphries generators or Dehn twists $\omega_{0}$, $\omega_{1}$,…., $\omega_{2g}$ along the loops depicted in Figure 1 (See [3, Theorem 4.14].) We consider an auxiliary Dehn twist $\omega_{2g+1}$ along the loop $c_{2g+1}$. Then the following relations hold:

[TABLE]

The element $\zeta_{0}$ is the hyperelliptic involution. If $1\leq k\leq m$, then

[TABLE]

since we obtain from (1) and (2)

[TABLE]

The second equation can be obtained in a similar way. Let

[TABLE]

Applying (5) we have $\omega_{i}\zeta=\zeta\omega_{i-1}$ for $i=2,...,2g+1$. Since

[TABLE]

we have $\omega_{1}\zeta=\zeta^{2}\eta^{-1}$, $\omega_{1}\zeta^{2}=\zeta^{2}\omega_{2g+1}$ and $\omega_{1}\zeta^{3}=\zeta^{2}\omega_{2g+1}\zeta=\zeta^{3}\omega_{2g}$. Continuing in this way we have for $i,j=1,...,2g+1$,

[TABLE]

where the index $k$ for $\omega_{k}$ is considered modulo $2g+2$. From this follows

[TABLE]

Lemma 2.1

(special cases of [3, Proposition 4.12]) * If $\eta=\omega_{1}\omega_{2}\cdots\omega_{2g}$ and $\xi=\omega_{1}^{2}\omega_{2}\cdots\omega_{2g}$, then $\eta^{2g+1}$ is a conjugate of $\zeta_{0}$ in (4) and $\xi^{2g}=\eta^{2g+1}$. Hence $\eta^{4g+2}=\xi^{4g}=1$ .*

Proof. Note that $\zeta_{0}=\zeta_{0}^{-1}$. From (7)

[TABLE]

A consequence of (5) is $\omega_{i+1}\eta=\eta\omega_{i}$ ($i=1,...,2g-1$). So we have

[TABLE]

From (7) we see that ${\cal MCG}_{g}$ is generated by three elements $\omega_{0}$, $\zeta$ and $\eta$. If $g=2$, then $\omega_{0}=\omega_{5}$. Therefore, we obtain a theorem by M. Korkmaz for $g=2$.

Corollary 2.1

(Korkmaz [6])* The mapping class group ${\cal MCG}_{2}$ is generated by $\zeta$ and $\eta$, where $\zeta^{6}=\eta^{10}=1$.*

2.2 Case of Genus $2$

The mapping class group ${\cal MCG}_{2}$ is generated by Humphries generators $\omega_{1}$, $\omega_{2}$, $\omega_{3}$, $\omega_{4}$ and $\omega_{5}$ with defining relations (1), (2) and

[TABLE]

See [1, p.184]. Let $\zeta_{0}$,…, $\zeta_{3}$ and $\zeta_{4}$ be as in the table of Section 1. From (8), the hyperelliptic involution $\zeta_{0}=\omega_{1}\omega_{2}\omega_{3}\omega_{4}\omega_{5}^{2}\omega_{4}\omega_{3}\omega_{2}\omega_{1}$ equals any conjugate of itself or its inverse. We proved in Lemma 2.1 that $\zeta_{3}^{4}=\zeta_{4}^{5}=\zeta_{0}$, and hence $\zeta_{3}^{8}=\zeta_{4}^{10}=1$. We shall show $\zeta_{2}^{6}=1$ in Lemma 3.1, but topologically this arises from $2$-chain relations [3, p.107] in one-holed tori on each side of the loop $d$ in Figure 2. We remark that S. Hirose studied in [4] presentations of periodic mapping classes on orientable closed surfaces of genus $\leq 4$ by Dehn twists by using topological and algebraic-geometric methods.

3 Proof of the main theorem

3.1 Abelian Groups

Our basic tools are elaborate applications of (6) for $g=2$

[TABLE]

and trivial equations

[TABLE]

Lemma 3.1

If $\zeta_{2}=\omega_{1}\omega_{2}\omega_{4}^{-1}\omega_{5}^{-1}$, then $\zeta_{2}^{3}=\zeta_{0}$. Hence $\zeta_{2}$ has order $6$.

Proof. The following equation deduced from (1), (9) and (11) implies $\zeta_{2}^{3}=\zeta_{0}$.

[TABLE]

Proofs of Theorem 1.1 for cyclic groups $G_{c}=\langle\zeta_{1}^{2}\rangle\cong\mathbb{Z}_{3}$, $G_{e}=\langle\zeta_{3}^{2}\rangle\cong\mathbb{Z}_{4}$, $G_{h}=\langle\zeta_{4}^{2}\rangle\cong\mathbb{Z}_{5}$, $G_{l}=\langle\zeta_{3}\rangle\cong\mathbb{Z}_{8}$ and $G_{o}=\langle\zeta_{4}\rangle\cong\mathbb{Z}_{10}$ are straightforward. Since $\zeta_{0}$ is in the center of $\mathcal{MCG}_{2}$, proofs for abelian groups $G_{f}=\langle\zeta_{0},\zeta_{1}^{3}\rangle\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and $G_{p}=\langle\zeta_{0},\zeta_{1}\rangle\cong\mathbb{Z}_{2}\times\mathbb{Z}_{6}$ are also easy.

There are isomorphic pairs of groups $(G_{a},G_{b})$ and $(G_{i},G_{k1})$. The types of orbifold for $G_{a}$ and $G_{k1}$ imply that they must contain the hyperelliptic involution. So $G_{a}=\langle\zeta_{0}\rangle\cong\mathbb{Z}_{2}$, $G_{b}=\langle\zeta_{1}^{3}\rangle\cong\mathbb{Z}_{2}$ and $G_{i}=\langle\zeta_{1}\rangle\cong\mathbb{Z}_{6}$.The equation (12) means that $G_{k1}=\langle\zeta_{2}\rangle\cong\langle\zeta_{1}^{2}\rangle\times\langle\zeta_{0}\rangle\cong\mathbb{Z}_{3}\times\mathbb{Z}_{2}\cong\mathbb{Z}_{6}$.

3.2 Non-abelian Groups

Groups with two or more generators require much effort. Assume that $x$ and $y$ generate a finite group of $\mathcal{MCG}_{2}$. We know each of them is a conjugate of a power of some $\zeta_{i}$, ($i=0,1,...,4$), but finding suitable conjugates of $x$ and $y$ so that their product has also a finite order is rather laborious.

3.2.1 Groups $G_{k2}$ and $G_{s}$

Let $a=\zeta_{1}^{3}$ and $b=\omega_{1}\omega_{2}\omega_{5}^{-1}\omega_{4}^{-1}=\omega_{4}\zeta_{2}\omega_{4}^{-1}$. Then $a^{2}=b^{6}=1$. It holds $(ba)^{2}=baba^{-1}=1$ because (9) yields

[TABLE]

So $a$ and $b$ generate $G_{s}\cong D_{6}$, and $a$ and $b^{2}$ generate $G_{k2}\cong D_{3}$.

3.2.2 Groups $G_{n}$, $G_{u}$

The group $G_{u}=\langle x,y:x^{2}=y^{8}=1,xyx^{-1}=y^{3}\rangle$ has the subgroup

[TABLE]

If $x=\zeta_{1}^{3}$ and $y=\omega_{1}\omega_{2}\omega_{4}\omega_{3}\omega_{2}=(\omega_{4}\omega_{2}^{-1}\omega_{1}^{-1})\zeta_{3}(\omega_{4}\omega_{2}^{-1}\omega_{1}^{-1})^{-1}$, then $x^{2}=1$, $y^{4}=\zeta_{0}$ and $y^{8}=1$. By using (8), (9) and (10),

[TABLE]

3.2.3 Groups $G_{r}$, $G_{w}$

Let $G_{w}=\mathbb{Z}_{2}\ltimes(\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{3})$ have the presentation as in (2.w). Since $[z,w]=1$, $u=zw$ generates $\mathbb{Z}_{6}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{3}$ and satisfies $u^{3}=z$ and $u^{4}=w$. Therefore $G_{w}$ can be written as

[TABLE]

This is an extension of the abelian group $G_{p}=\langle y,u:y^{2}=u^{6}=[y,u]=1\rangle$. By letting $t=xu^{3}$ and $w=u^{4}$, we find the subgroup

[TABLE]

of $G_{w}$. Let $\zeta_{5}=\omega_{1}\omega_{2}\omega_{1}\omega_{4}^{-1}\omega_{5}^{-1}\omega_{4}^{-1}$. By using (1), (2), (9) and (11) we have

[TABLE]

Thus we obtain

[TABLE]

and also

[TABLE]

Let $x=\zeta_{5}\zeta_{1}^{3}=(\omega_{1}\omega_{2}\omega_{1})\zeta_{1}^{3}(\omega_{1}\omega_{2}\omega_{1})^{-1}$, $y=\zeta_{0}$ and $u=\zeta_{1}$. Then $y^{2}=u^{6}=[x,y]=[y,u]=1$. The relations $x^{2}=1$ and $xux^{-1}=u^{-1}y$ are equivalent to $(\zeta_{5}\zeta_{1}^{3})^{2}=1$ and $(\zeta_{5}\zeta_{1}^{4})^{2}=\zeta_{0}$, which follow from (14).

3.2.4 Groups $G_{m}$, $G_{x}$, $G_{aa}$

$SL_{2}(3)=\langle x,y:x^{3}=y^{4}=1,xy^{2}=y^{2}x,(xy)^{3}=1\rangle$, where

[TABLE]

$GL_{2}(3)$ is obtained by adding to $SL_{2}(3)$ the matrix

[TABLE]

which satisfies $u^{8}=1$, $uy^{2}=y^{2}u$ and

[TABLE]

By letting $v=x^{-1}yx$, we find the subgroup $G_{m}$ of $SL_{2}(3)$ presented by

[TABLE]

Now, we represent $x$, $y$ and $u$ by $\omega_{1}$,…, $\omega_{5}$. By using (1) and (9)

[TABLE]

Hence $(\zeta_{2}\zeta_{3}^{-2})^{3}=1$. Let $x=\zeta_{2}\zeta_{0}=\zeta_{2}^{4}$ and $y=\zeta_{3}^{2}$. Then we have $x^{3}=y^{4}=1$, $y^{2}=\zeta_{0}$, $(xy)^{3}=(\zeta_{2}\zeta_{3}^{-2})^{3}=1$ and $xy^{2}=y^{2}x$. Let

[TABLE]

and $u=\omega_{1}c\omega_{1}^{-1}$. Then $u^{8}=1$. Since $(x,y,u)=(\omega_{1}a\zeta_{0}\omega_{1}^{-1},\omega_{1}b^{2}\omega_{1}^{-1},\omega_{1}c\omega_{1}^{-1})$, the relations $uxu^{-1}=y^{-1}x^{-1}y$ and $b^{2}ca(b^{2}c)^{-1}=a^{-1}$ are equivalent. By using (9) we have $\omega_{1}\zeta_{4}=\zeta_{4}\omega_{4}^{-1}\omega_{3}^{-1}\omega_{2}^{-1}\zeta_{4}=\zeta_{4}^{2}\omega_{3}^{-1}\omega_{2}^{-1}\omega_{1}^{-1}$, and then

[TABLE]

On the other hand, from $\omega_{2}\omega_{1}\omega_{4}^{-1}\omega_{5}^{-1}\zeta_{1}^{3}=\omega_{2}\omega_{1}\zeta_{1}^{3}\omega_{1}^{-1}\omega_{2}^{-1}$ we have

[TABLE]

Since $\zeta_{1}^{3}=\zeta_{1}^{-3}$ this means $b^{2}ca(b^{2}c)^{-1}=a^{-1}$. The relations $uyu^{-1}=x^{-1}yx$ and $(ac)b^{2}(ac)^{-1}=b^{2}$ are equivalent. The last relation easily follows from

[TABLE]

Finally we show the first relation $u^{2}=xyx^{-1}y^{2}$ in (15), which is equivalent to $c^{2}=ab^{2}a^{-1}b^{4}$. Since $b^{4}=b^{-4}=\zeta_{0}$, $ab^{2}a^{-1}b^{4}=ab^{-2}a^{-1}=(ab^{-1}a^{-1})^{2}$. We obtain $c^{2}=(ab^{-1}a^{-1})^{2}$ from

[TABLE]