On the spectrum of genera of quotients of the Hermitian curve

TL;DR

This paper studies the genera of quotient curves derived from the Hermitian curve over finite fields, providing new genus values and geometric models, especially for quotients by certain subgroups of the automorphism group.

Contribution

It offers a detailed geometric and group-theoretical analysis of the automorphism subgroup fixing a pole-polar pair, and computes new genus values for quotient curves beyond known cases.

Findings

New genus values for quotient curves of the Hermitian curve.

A geometric model for quotients when the group is cyclic of order p·d.

Extension of the spectrum of genera for maximal curves.

Abstract

We investigate the genera of quotient curves $\mathcal H_q/G$ of the $\mathbb F_{q^2}$-maximal Hermitian curve $\mathcal H_q$, where $G$ is contained in the maximal subgroup $\mathcal M_q\leq{\rm Aut}(\mathcal H_q)$ fixing a pole-polar pair $(P,\ell)$ with respect to the unitary polarity associated with $\mathcal H_q$. To this aim, a geometric and group-theoretical description of $\mathcal M_q$ is given. The genera of some other quotients $\mathcal H_q/G$ with $G\not\leq\mathcal M_q$ are also computed. Thus we obtain new values in the spectrum of genera of $\mathbb F_{q^2}$-maximal curves. A plane model for $\mathcal H_q/G$ is obtained when $G$ is cyclic of order $p\cdot d$, with $d$ a divisor of $q+1$.

Tables

Table 1. Table 1. Values of i ​ ( σ ) 𝑖 𝜎 i(\sigma) for σ ∈ S ​ L ​ ( 2 , 5 ) 𝜎 𝑆 𝐿 2 5 \sigma\in SL(2,5)

Order of $\sigma$	number of elements in $SL(2,5)$	$q{\equiv }_{12}1$	$q{\equiv }_{12}7$	$q{\equiv }_{12}5$	$q{\equiv }_{12}11$
2	1	q+1	q+1	q+1	q+1
3	20	2	2	0	0
4	30	2	0	2	0
5	24	2	2	2	2
6	20	2	2	0	0
10	24	2	2	2	2
		$i\left(\sigma \right)$	$i\left(\sigma \right)$	$i\left(\sigma \right)$	$i\left(\sigma \right)$

Table 2. Table 2. Values of i ​ ( σ ) 𝑖 𝜎 i(\sigma) for σ ∈ S ​ L ​ ( 2 , 3 ) 𝜎 𝑆 𝐿 2 3 \sigma\in SL(2,3)

Order of $\sigma$	number of elements in $SL(2,3)$	$q{\equiv }_{12}1$	$q{\equiv }_{12}7$	$q{\equiv }_{12}5$	$q{\equiv }_{12}11$
2	1	q+1	q+1	q+1	q+1
3	8	2	2	0	0
4	6	2	0	2	0
6	8	2	2	0	0
		$i\left(\sigma \right)$	$i\left(\sigma \right)$	$i\left(\sigma \right)$	$i\left(\sigma \right)$

Table 3. Table 3. Values of i ​ ( σ ) 𝑖 𝜎 i(\sigma) for σ ∈ K 𝜎 𝐾 \sigma\in K

$o\left(\sigma \right)$	$\#\{\gamma \mid o\left(\gamma \right)=o\left(\sigma \right)\}$	$q{\equiv }_{24}1,13$	$q{\equiv }_{24}5,17$	$q{\equiv }_{24}7,19$	$q{\equiv }_{24}11,23$
2	1	q+1	q+1	q+1	q+1
3	8	2	0	2	0
4	18	2	2	0	0
6	8	2	0	2	0
8	12	2	2	0	0

Table 4. Table 4. Quotient curves ℋ 2 / G subscript ℋ 2 𝐺 {\mathcal{H}}_{2}/G ,   G ≤ PGU ​ ( 3 , 2 ) 𝐺 PGU 3 2 G\leq{\rm PGU}(3,2)

$g$	$\left|G\right|$	structure of $G$
1	1	trivial group.
0	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (C).

Table 5. Table 5. Quotient curves ℋ 3 / G subscript ℋ 3 𝐺 {\mathcal{H}}_{3}/G ,   G ≤ PGU ​ ( 3 , 3 ) 𝐺 PGU 3 3 G\leq{\rm PGU}(3,3)

$g$	$\left|G\right|$	structure of $G$
3	1	trivial group.
1	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
0	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (C).

Table 6. Table 6. Quotient curves ℋ 4 / G subscript ℋ 4 𝐺 {\mathcal{H}}_{4}/G ,   G ≤ PGU ​ ( 3 , 4 ) 𝐺 PGU 3 4 G\leq{\rm PGU}(3,4)

$g$	$\left|G\right|$	structure of $G$
6	1	trivial group.
2	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B2).
1	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (D).
0	5	$G=C_5=⟨\sigma ⟩$, $\sigma$ of type (A).

Table 7. Table 7. Quotient curves ℋ 5 / G subscript ℋ 5 𝐺 {\mathcal{H}}_{5}/G ,   G ≤ PGU ​ ( 3 , 5 ) 𝐺 PGU 3 5 G\leq{\rm PGU}(3,5)

$g$	$\left|G\right|$	structure of $G$
10	1	trivial group.
4	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
3	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B3).
2	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (B2).
1	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (A).
0	5	$G=C_5=⟨\sigma ⟩$, $\sigma$ of type (C).

Table 8. Table 8. Quotient curves ℋ 7 / G subscript ℋ 7 𝐺 {\mathcal{H}}_{7}/G ,   G ≤ PGU ​ ( 3 , 7 ) 𝐺 PGU 3 7 G\leq{\rm PGU}(3,7)

$g$	$\left|G\right|$	structure of $G$
21	1	trivial group.
9	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
7	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B2).
5	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (B1).
3	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (A).
2	6	$G=Sym\left(3\right)=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
1	8	$G=C_4\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
0	7	$G=C_7=⟨\sigma ⟩$, $\sigma$ of type (C).

Table 9. Table 9. Quotient curves ℋ 8 / G subscript ℋ 8 𝐺 {\mathcal{H}}_{8}/G ,   G ≤ PGU ​ ( 3 , 8 ) 𝐺 PGU 3 8 G\leq{\rm PGU}(3,8)

$g$	$\left|G\right|$	structure of $G$
28	1	trivial group.
12	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (C).
10	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B1).
9	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B3).
7	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (A).
6	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (D).
4	7	$G=C_7=⟨\sigma ⟩$, $\sigma$ of type (B2).
3	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (E).
2	8	$G=C_4\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (D), $\beta$ of type $\left(C\right)$.
1	19	$G=C_{1}9=⟨\sigma ⟩$, $\sigma$ of type (B3).
0	9	$G=C_9=⟨\sigma ⟩$, $\sigma$ of type (A).

Table 10. Table 10. Quotient curves ℋ 9 / G subscript ℋ 9 𝐺 {\mathcal{H}}_{9}/G ,   G ≤ PGU ​ ( 3 , 9 ) 𝐺 PGU 3 9 G\leq{\rm PGU}(3,9)

$g$	$\left|G\right|$	structure of $G$
36	1	trivial group.
16	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
12	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (D).
9	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (C).
8	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (B2).
6	4	$G=C_2\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
4	5	$G=C_5=⟨\sigma ⟩$, $\sigma$ of type (A).
3	9	$G=C_3\times C_3=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (C), $\beta$ of type (D).
2	8	$G=D_8=C_4⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type $\left(A\right)$.
1	15	$G=C_{15}=⟨\sigma ⟩$, $\sigma$ of type (E).
0	10	$G=C_{10}=⟨\sigma ⟩$, $\sigma$ of type (A).

Table 11. Table 11. Quotient curves ℋ 11 / G subscript ℋ 11 𝐺 {\mathcal{H}}_{11}/G ,   G ≤ PGU ​ ( 3 , 11 ) 𝐺 PGU 3 11 G\leq{\rm PGU}(3,11)

$g$	$\left|G\right|$	structure of $G$
55	1	trivial group.
25	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
19	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B1).
18	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B3).
15	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (A).
13	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (B1).
11	5	$G=C_5=⟨\sigma ⟩$, $\sigma$ of type (B2).
10	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (A).
9	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (B1).
7	8	$G=Q_8$.
5	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (A).
4	8	$G=D_8=C_4⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (A).
3	10	$D_{10}=C_5⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
2	15	$G=C_{15}=⟨\sigma ⟩$, $\sigma$ of type (B2).
1	37	$G=C_{37}=⟨\sigma ⟩$, $\sigma$ of type (B3).
0	12	$G=C_{12}=⟨\sigma ⟩$, $\sigma$ of type (A).

Table 12. Table 12. Quotient curves ℋ 13 / G subscript ℋ 13 𝐺 {\mathcal{H}}_{13}/G ,   G ≤ PGU ​ ( 3 , 13 ) 𝐺 PGU 3 13 G\leq{\rm PGU}(3,13)

$g$	$\left|G\right|$	structure of $G$
78	1	trivial group.
36	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
26	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B2).
18	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (B2).
15	4	$G=C_2\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
12	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (B2).
10	6	$G=S_3$.
9	8	$G=Q_8$.
6	8	$G=D_8=C_4⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
5	12	$G=A_4$.
4	21	$G=C_7⋊C_3=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (B2).
3	14	$G=D_{14}=C_7⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (A).
2	21	$G=C_{21}=⟨\sigma ⟩$, $\sigma$ of type (B2).
0	14	$G=C_{14}=⟨\sigma ⟩$, $\sigma$ of type (A).

Table 13. Table 13. Quotient curves ℋ 16 / G subscript ℋ 16 𝐺 {\mathcal{H}}_{16}/G ,   G ≤ PGU ​ ( 3 , 16 ) 𝐺 PGU 3 16 G\leq{\rm PGU}(3,16)

$g$	$\left|G\right|$	structure of $G$
120	1	trivial group.
56	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (C).
40	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B2).
28	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (D).
24	4	$G=C_2\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
16	6	$G=Sym\left(3\right)=C_3⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2) and $\beta$ of type (C).
12	8	$G=C_4\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (D) and $\beta$ of type (C).
8	8	$G=C_2\times C_2\times C_2$.
6	16	$G=C_4\times C_4=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (D), $\beta$ of type (D).
4	16	$G=C_2\times C_2\times C_4$.
2	32	$G=C_2\times C_4\times C_4$.
1	64	$G=C_4\times C_4\times C_4$.
0	17	$G=C_{17}=⟨\sigma ⟩$, $\sigma$ of type (A).

Table 14. Table 14. Quotient curves ℋ 17 / G subscript ℋ 17 𝐺 {\mathcal{H}}_{17}/G ,   G ≤ PGU ​ ( 3 , 17 ) 𝐺 PGU 3 17 G\leq{\rm PGU}(3,17)

$g$	$\left|G\right|$	structure of $G$
136	1	trivial group.
64	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
46	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B1).
45	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B3).
40	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (A).
32	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (B2).
28	4	$G=C_2\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
22	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (B1).
19	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (B1).
16	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (A).
14	9	$G=C_9=⟨\sigma ⟩$, $\sigma$ of type (B1).
12	8	$G=D_8=C_4⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
11	8	$G=Di{c}_{12}$.
10	12	$G=A_4$.
8	9	$G=C_9=⟨\sigma ⟩$, $\sigma$ of type (A).
7	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (A).
6	18	$G=C_{18}=⟨\sigma ⟩$, $\sigma$ of type (B1).
5	18	$G=C_6\times C_3=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (B1).
4	18	$G=C_6\times C_3=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
3	18	$G=Sym\left(3\right)\times C_3$.
2	18	$G=C_3\times C_3⋊C_2=⟨\alpha ⟩\times ⟨\beta ⟩\times ⟨\gamma ⟩$, $\alpha$, $\beta$, $\gamma$ of type (A).
1	91	$G=C_{91}=⟨\sigma ⟩$, $\sigma$ of type (B3).
0	18	$G=C_{18}=⟨\sigma ⟩$, $\sigma$ of type (A).

Table 15. Table 15. Quotient curves ℋ 19 / G subscript ℋ 19 𝐺 {\mathcal{H}}_{19}/G ,   G ≤ PGU ​ ( 3 , 19 ) 𝐺 PGU 3 19 G\leq{\rm PGU}(3,19)

$g$	$\left|G\right|$	structure of $G$
171	1	trivial group.
81	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
57	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B2).
41	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (B1).
36	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (A).
35	5	$G=C_5=⟨\sigma ⟩$, $\sigma$ of type (B1).
27	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (A).
24	6	$G=Sym\left(3\right)=C_3⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
21	6	$G=Q_8$.
19	9	$G=C_9=⟨\sigma ⟩$, $\sigma$ of type (B2).
18	8	$G=C_8=⟨\sigma ⟩$, $\sigma$ of type (B2).
17	10	$G=C_{10}=⟨\sigma ⟩$, $\sigma$ of type (B1).
16	8	$G=D_8=C_4⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (A).
14	12	$G=Di{c}_{12}$.
13	10	$G=C_{10}=⟨\sigma ⟩$, $\sigma$ of type (B1).
12	12	$G=C_{12}=⟨\sigma ⟩$, $\sigma$ of type (B2).
9	10	$G=C_{10}=⟨\sigma ⟩$, $\sigma$ of type (A).
8	21	$G=C_7⋊C_3=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B3), $\beta$ of type (B2).
7	24	$G=SL(2,3)$.
6	24	$G=C_3⋊C_8=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (B2).
5	18	$G=D_9=C_9⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
4	24	$G=Sym\left(3\right)\times C_4$.
3	30	$G=C_{30}=⟨\sigma ⟩$, $\sigma$ of type (B2).
2	18	$G=SG(32,11)$.
1	141	$G=C_{49}⋊C_3=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B3), $\beta$ of type (B2).
0	20	$G=C_{20}=⟨\sigma ⟩$, $\sigma$ of type (A).

Table 16. Table 16. Quotient curves ℋ 23 / G subscript ℋ 23 𝐺 {\mathcal{H}}_{23}/G ,   G ≤ PGU ​ ( 3 , 23 ) 𝐺 PGU 3 23 G\leq{\rm PGU}(3,23)

$g$	$\left|G\right|$	structure of $G$
253	1	trivial group.
121	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
85	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B1).
84	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B3).
77	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (A).
61	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (B1).
55	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (A).
41	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (B1).
37	6	$G=Sym\left(3\right)=C_3⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (A).
33	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (A).
31	8	$G=Q_8$.
28	8	$G=C_8=⟨\sigma ⟩$, $\sigma$ of type (B1).
25	8	$G=C_4\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
23	11	$G=C_{11}=⟨\sigma ⟩$, $\sigma$ of type (B2).
22	8	$G=C_8=⟨\sigma ⟩$, $\sigma$ of type (A).
21	12	$G=C_{12}=⟨\sigma ⟩$, $\sigma$ of type (B1).
19	12	$G=C_{12}=⟨\sigma ⟩$, $\sigma$ of type (B1).
17	12	$G=C_{12}=⟨\sigma ⟩$, $\sigma$ of type (B1).
16	16	$G=Q_{16}$.
15	12	$G=C_6\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
13	16	$G=C_8\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (A).
11	12	$G=C_{12}=⟨\sigma ⟩$, $\sigma$ of type (A).
10	16	$G=C_4\times C_4=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
9	18	$G=C_6\times C_3=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
8	24	$G=C_{24}=⟨\sigma ⟩$, $\sigma$ of type (B1).
7	33	$G=C_{33}=⟨\sigma ⟩$, $\sigma$ of type (B2).
6	22	$G=C_{11}⋊C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
5	18	$G=C_3\times C_3⋊C_2=⟨\alpha ⟩\times ⟨\beta ⟩⋊⟨\gamma ⟩$, $\alpha$, $\beta$ of type (A), $\gamma$ of type (A) .
4	32	$G=C_8\times C_4=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
3	24	$G=D_8\times C_3=\left(C_4⋊C_2\right)\times C_3=\left(⟨\alpha ⟩⋊⟨\beta ⟩\right)\times ⟨\gamma ⟩$, $\alpha$ of type (B1), $\beta$, $\gamma$ of type (A).
2	88	$G=C_{88}=⟨\sigma ⟩$, $\sigma$ of type (B2).
1	169	$G=C_{169}=⟨\sigma ⟩$, $\sigma$ of type (B3).
0	24	$G=C_{24}=⟨\sigma ⟩$, $\sigma$ of type (A).

Table 17. Table 17. Quotient curves ℋ 25 / G subscript ℋ 25 𝐺 {\mathcal{H}}_{25}/G ,   G ≤ PGU ​ ( 3 , 25 ) 𝐺 PGU 3 25 G\leq{\rm PGU}(3,25)

$g$	$\left|G\right|$	structure of $G$
300	1	trivial group.
144	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
100	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B2).
72	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (B2).
66	4	$G=C_2\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (A), $\beta$ of type (A).
60	5	$G=C_5=⟨\sigma ⟩$, $\sigma$ of type (D).
50	5	$G=C_5=⟨\sigma ⟩$, $\sigma$ of type (C).
48	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (B2).
44	6	$G=Sym\left(3\right)=C_3⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
36	8	$G=Q_8$.
30	8	$G=D_8=C_4⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
24	10	$G=C_{10}=⟨\sigma ⟩$, $\sigma$ of type (E).
22	12	$G=C_6\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
18	12	$G=D_{12}=C_6⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
12	13	$G=C_{13}=⟨\sigma ⟩$, $\sigma$ of type (A).
10	25	$G=C_5\times C_5=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (C), $\beta$ of type (D).
8	39	$G=C_{13}⋊C_3=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (B2).
6	24	$G=D_{24}=C_{12}⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
4	39	$G=C_{39}=⟨\sigma ⟩$, $\sigma$ of type (B2).
3	52	$G=C_{13}⋊C_4=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (B2).
2	125	$G=C_5\times C_5\times C_5=⟨\alpha ⟩\times ⟨\beta ⟩\times ⟨\gamma ⟩$, $\alpha$ of type (C), $\beta$, $\gamma$ of type (D).
0	26	$G=C_{26}=⟨\sigma ⟩$, $\sigma$ of type (A).

Table 18. Table 18. Quotient curves ℋ 27 / G subscript ℋ 27 𝐺 {\mathcal{H}}_{27}/G , G ≤ PGU ​ ( 3 , 27 ) 𝐺 PGU 3 27 G\leq{\rm PGU}(3,27)

$g$	$\left|G\right|$	structure of $G$
351	1	trivial group.
169	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
117	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (D).
108	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (C).
85	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (B1).
78	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (A).
52	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (E).
		$G=Sym\left(3\right)=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (C), $\beta$ of type (A).
51	7	$G=C_7=⟨\sigma ⟩$, $\sigma$ of type (B1).
43	8	$G=Q_8$ quaternion group, 1 element of type (A), 6 elements of type (B1).
39	7	$G=C_7=⟨\sigma ⟩$, $\sigma$ of type (A).
27	9	$G=C_3\times C_3=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ and $\beta$ of type (C).
26	12	$G=Alt\left(4\right)$, involutions of type (A), other elements of type (D).
25	14	$G=C_{14}=⟨\sigma ⟩$, $\sigma$ of type (B1).
24	12	$G=C_{12}=⟨\sigma ⟩$, $\sigma$ of type (E).
19	14	$G=C_{14}=⟨\sigma ⟩$, $\sigma$ of type (B1), ${\sigma }^2$ of type (A).
18	16	$G=M_{16}$, 5 elements of type (A),
		2 elements of type (B1), 8 elements of type (B2).
18	19	$G=C_{19}=⟨\sigma ⟩$, $\sigma$ of type (B3).
17	21	$G=C_7⋊C_3=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (B2).
16	18	$G=C_3\times \left(C_3⋊C_2\right)=⟨\alpha ⟩\times \left(⟨\beta ⟩⋊⟨\gamma ⟩\right)$,
		$\alpha$ of type (C), $\beta$ of type (D), $\gamma$ of type (A).
15	16	$G=D_8\circ C_4=\left(⟨\alpha ⟩⋊⟨\beta ⟩\right)\circ ⟨\gamma ⟩$, $\alpha$ of type (B1), $\beta$ and $\gamma$ of type (A).
13	14	$G=C_{14}=⟨\sigma ⟩$, $\sigma$ of type (A).
12	21	$G=C_{21}=⟨\sigma ⟩$, $\sigma$ of type (E).
10	24	$G\cong \mathrm{SL}(2,3)$, 1 element of type (A), 6 elements of type (B1),
		8 elements of type (C), 8 elements of type (E).
9	37	$G=C_{37}=⟨\sigma ⟩$, $\sigma$ of type (B3).
7	26	$G=C_{13}⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
6	32	$G=C_4\wr C_2=⟨\alpha ⟩\wr ⟨\beta ⟩$ wreath product,
		13 elements of type (A), 10 elements of type (B1), 8 elements of type (B2).
5	48	$G=\left(C_4\times C_4\right)⋊C_3=\left(⟨\alpha ⟩\times ⟨\beta ⟩\right)⋊⟨\gamma ⟩$, $\alpha$ and $\beta$ of type (A), $\gamma$ of type (D).
4	48	$G=\left(D_8\circ C_4\right)⋊⟨\sigma ⟩$, $\sigma$ of type (C).
3	49	$G=C_7\times C_7=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ and $\beta$ of type (A).
1	126	$G=C_{14}\times C_3\times C_3=⟨\alpha ⟩\times ⟨\beta ⟩\times ⟨\gamma ⟩$, $\alpha$ of type (A), $\beta$, $\gamma$ of type (C).
0	28	$G=C_{28}=⟨\sigma ⟩$, $\sigma$ of type (A).

Table 19. Table 19. Quotient curves ℋ 29 / G subscript ℋ 29 𝐺 {\mathcal{H}}_{29}/G , G ≤ PGU ​ ( 3 , 29 ) 𝐺 PGU 3 29 G\leq{\rm PGU}(3,29)

$g$	$\left|G\right|$	structure of $G$
406	1	trivial group.
196	2	$G=C_2=⟨\sigma ⟩$, $\sigma$ of type (A).
136	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B1).
135	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (B3).
126	3	$G=C_3=⟨\sigma ⟩$, $\sigma$ of type (A).
98	4	$G=C_4=⟨\sigma ⟩$, $\sigma$ of type (B2).
91	4	$G=C_2\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ and $\beta$ of type (A).
82	5	$G=C_5=⟨\sigma ⟩$, $\sigma$ of type (B1).
70	5	$G=C_5=⟨\sigma ⟩$, $\sigma$ of type (A).
66	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (B1).
61	6	$G=Sym\left(3\right)=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (A).
58	7	$G=C_7=⟨\sigma ⟩$, $\sigma$ of type (B2).
56	6	$G=C_6=⟨\sigma ⟩$, $\sigma$ of type (A).
49	8	$G=Q_8=⟨\alpha ,\beta ⟩$, $\alpha$ and $\beta$ of type (B2).
45	9	$G=C_3\times C_3=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (B3), $\beta$ of type (B1).
42	8	$G=D_8=C_4⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
40	10	$G=C_{10}=⟨\sigma ⟩$, $\sigma$ of type (B1).
36	9	$G=C_3\times C_3=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ and $\beta$ of type (A).
34	10	$G=C_5⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (A).
33	12	$G=Di{c}_{12}=⟨\alpha ,\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (B2).
31	12	$G=Alt\left(4\right)=⟨\alpha ⟩\times ⟨\beta ⟩⋊⟨\gamma ⟩$, $\alpha$ and $\beta$ of type (A), $\gamma$ of type (B1).
28	10	$G=C_{10}=⟨\sigma ⟩$, $\sigma$ of type (A).
26	12	$G=D_{12}=C_6⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (A).
24	15	$G=C_{15}=⟨\sigma ⟩$, $\sigma$ of type (B1).
22	14	$G=D_{14}=C_7⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
21	16	$G=S{D}_{16}$, semidihedral group (Sylow $2$-subgroup of $\mathrm{PGU}(3,29)$).
20	20	$D=Di{c}_{20}$.
19	20	$G=C_{10}\times C_2=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (A).
18	21	$G=C_{21}=⟨\sigma ⟩$, $\sigma$ of type (B2).
17	24	$G=SL(2,3)$.
16	18	$G=C_6\times C_3=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ and $\beta$ of type (A).
14	15	$G=C_{15}=⟨\sigma ⟩$, $\sigma$ of type (A).
13	20	$G=D_{20}=C_{10}⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (A).
12	24	$G=D_8\times C_3=\left(⟨\alpha ⟩⋊⟨\beta ⟩\right)\times ⟨\gamma ⟩$, $\alpha$ of type (B2), $\beta$ and $\gamma$ of type (A).
11	30	$G=D_{30}=C_{15}⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B1), $\beta$ of type (A).
10	25	$G=C_5\times C_5=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$, $\beta$ of type (A).
8	36	$G=\left(C_3\times C_3\right)⋊C_4=\left(⟨\alpha ⟩\times ⟨\beta ⟩\right)⋊⟨\gamma ⟩$, $\alpha ,\beta$ of type (A), $\gamma$ of type (B2).
7	24	$G=D_{24}=C_{12}⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
6	36	$G=C_3\times C_3\times C_2\times C_2.$
5	48	$G=S{D}_{16}⋊C_3$.
4	45	$G=C_{15}\times C_3=⟨\alpha ⟩\times ⟨\beta ⟩$, $\alpha$ and $\beta$ of type (A).
3	120	$G=Q_8⋊C_{15}=Q_8⋊⟨\alpha ⟩$,$\alpha$ of type (B1).
2	42	$G=D_{42}=C_{21}⋊C_2=⟨\alpha ⟩⋊⟨\beta ⟩$, $\alpha$ of type (B2), $\beta$ of type (A).
1	271	$G=C_{271}=⟨\sigma ⟩$, $\sigma$ of type (B3).
0	30	$G=C_{30}=⟨\sigma ⟩$, $\sigma$ of type (A).

Equations

$$X^{q+1}+Y^{q+1}+Z^{q+1}=0;$$

$$X^{q}Z+X{Z}^q-Y^{q+1}=0;$$

$$X{Y}^q-Y{X}^q+\omega Z^{q+1}=0,$$

$$X{Y}^q+Y{Z}^q+\omega Z{X}^q=0.$$

$$k:{\mathbb{P}}^2(\mathbb{K})\to {\mathbb{P}}^2(\mathbb{K}),(x:y:z)\mapsto (ax+y+a^{q^2+1}z:a^{q^2+1}x+ay+z:x+a^{q^2+1}y+az),$$

$$d_{\pi }=\left(\begin{array}{ccc}w& 0& 0\\ 0& w^{-1}& 0\\ 0& 0& 1\end{array}\right),$$

$$i(\sigma )={\sum }_{P\in {\mathcal{H}}_q({\bar{\mathbb{F}}}_q)}{v}_P(\sigma (t)-t),$$

$$(X,Y,Z)\mapsto (aX+bY,cX+dY,Z),$$

$$a{c}^q-a^{q}c=0,b{d}^q-b^{q}d=0,b{c}^q-a^{q}d=-1,a{d}^q-b^{q}c=1.$$

$$\tau :(X,Y,Z)\mapsto (-X,-Y,Z).$$

$${\mathcal{M}_{q}}=C_{{\rm PGU}(3,q)}(\tau)=\Bigg{\{}\begin{pmatrix}a&b&0\\ c&d&0\\ 0&0&1\end{pmatrix}\mid ac^{q}-a^{q}c=0,\ bd^{q}-b^{q}d=0,\ bc^{q}-a^{q}d=-1,\ ad^{q}-b^{q}c=1\Bigg{\}},$$

$$H=({\mathcal{M}_{q}})^{\prime}=\Bigg{\{}\begin{pmatrix}a&b&0\\ c&d&0\\ 0&0&1\end{pmatrix}\in{\mathcal{M}_{q}}\mid a,b,c,d\in\mathbb{F}_{q}\ and\ ad-bc=1\Bigg{\}}\cong SL(2,q),$$

$$Z({\mathcal{M}_{q}})=\Bigg{\{}\begin{pmatrix}a&0&0\\ 0&a&0\\ 0&0&1\end{pmatrix}\in{\mathcal{M}_{q}}\mid a^{q+1}=1\Bigg{\}}\cong C_{q+1}.$$

$${\mathcal{M}}_q=H⋊Z(C_{\mathrm{PGU}(3,q)}({\tau }^{\prime }))\cong SL(2,q)⋊C_{q+1},$$

$${\mathcal{M}}_q=H⋊⟨\alpha ⟩,$$

$$\alpha =\left(\begin{array}{ccc}0& e^{-1}& 0\\ -e^q& 1+e^{q+1}& 0\\ 0& 0& 1\end{array}\right).$$

$$\sigma =\left(\begin{array}{ccc}a& 0& 0\\ 0& a^{-1}& 0\\ 0& 0& 1\end{array}\right),$$

$$\varphi :H\cong SL(2,q)\to PSL(2,q).$$

$$\bar{g}=\frac{(q-p^k)(q+1-\gcd (d,2))}{2d{p}^k}$$

$$\bar{g}=\frac{q^2-22q+117-48[h+2{]}_{mod4}}{240}.$$

$$i(\sigma )=\{\begin{array}{l}q+1,ifo(\sigma )=2;\\ q+2,ifo(\sigma )=3;\\ 2,ifo(\sigma )\in \{5,4,10\};\\ 1,ifo(\sigma )=6.\end{array}$$

$$i(\sigma )=\{\begin{array}{l}q+1,ifo(\sigma )=2;\\ q+2,ifo(\sigma )=3;\\ 0,ifo(\sigma )\in \{5,10\};\\ 2,ifo(\sigma )=4;\\ 1,ifo(\sigma )=6.\end{array}$$

$$\bar{g}=\{\begin{array}{ll}\frac{(q-\bar{q})(q-{\bar{q}}^2+\bar{q}-1)}{2\bar{q}({\bar{q}}^2-1)},& \text{if}q={\bar{q}}^{2r+1},r\ge 0;\text{vspace}\ast 2mm\\ \frac{(q-1)(q-{\bar{q}}^2)}{2\bar{q}({\bar{q}}^2-1)},& \text{if}q={\bar{q}}^{2r},r\ge 1.\end{array}$$

$$(q+1)(q-2)=$$

$$\bar{q}(\bar{q}+1)(\bar{q}-1)(2\bar{g}-2)+(q+1)+2\frac{\bar{q}(\bar{q}+1)(\bar{q}-3)}{2}+0\frac{\bar{q}(\bar{q}-1{)}^2}{2}+(\bar{q}-1)(\bar{q}+1)(q+2)+({\bar{q}}^2-1),$$

$$(q+1)(q-2)=$$

$$\bar{q}(\bar{q}+1)(\bar{q}-1)(2\bar{g}-2)+(q+1)+2\frac{\bar{q}(\bar{q}+1)(\bar{q}-3)}{2}+2\frac{\bar{q}(\bar{q}-1{)}^2}{2}+(\bar{q}-1)(\bar{q}+1)(q+2)+({\bar{q}}^2-1).$$

$$d_{\pi }=\left(\begin{array}{ccc}w& 0& 0\\ 0& w^{-1}& 0\\ 0& 0& 1\end{array}\right),$$

$$\bar{g}=\frac{(q-{\bar{q}}^2)(q-1)}{4\bar{q}({\bar{q}}^2-1)}.$$

$$(q+1)(q-2)=2\bar{q}(\bar{q}+1)(\bar{q}-1)(2\bar{g}-2)+(q+1)+2\frac{\bar{q}(\bar{q}+1)(\bar{q}-3)}{2}$$

Full text

Keywords: Hermitian curve, Unitary groups, quotient curves, maximal curves

2000 MSC: 11G20

1. Introduction

Let $q$ be a power of a prime $p$, $\mathbb{F}_{q^{2}}$ be the finite field with $q^{2}$ elements, and $\mathcal{X}$ be an $\mathbb{F}_{q^{2}}$-rational curve, i.e. a projective, absolutely irreducible, non-singular algebraic curve defined over $\mathbb{F}_{q^{2}}$. The curve $\mathcal{X}$ is called $\mathbb{F}_{q^{2}}$-maximal if the number $|\mathcal{X}(\mathbb{F}_{q^{2}})|$ of its ${\mathbb{F}_{q^{2}}}$-rational points attains the Hasse-Weil upper bound $q^{2}+1+2gq$, where $g$ is the genus of $\mathcal{X}$. Maximal curves have interesting properties and have also been investigated for their applications in Coding Theory. Surveys on maximal curves are found in [17, Chapter 10].

The most important example of an ${\mathbb{F}_{q^{2}}}$-maximal curve is the Hermitian curve ${\mathcal{H}}_{q}$, defined as any ${\mathbb{F}_{q^{2}}}$-rational curve projectively equivalent to the plane curve with Fermat equation $X^{q+1}+Y^{q+1}+Z^{q+1}=0$. For fixed $q$, ${\mathcal{H}}_{q}$ has the largest possible genus $q(q-1)/2$ that an ${\mathbb{F}_{q^{2}}}$-maximal curve can have. The full automorphism group ${\rm Aut}({\mathcal{H}}_{q})$ is isomorphic to $PGU(3,q)$, the group of projectivities of $PG(2,q^{2})$ commuting with the unitary polarity associated with ${\mathcal{H}}_{q}$.

By a result commonly attributed to Serre, any ${\mathbb{F}_{q^{2}}}$-rational curve which is ${\mathbb{F}_{q^{2}}}$-covered by an ${\mathbb{F}_{q^{2}}}$-maximal curve is also ${\mathbb{F}_{q^{2}}}$-maximal. In particular, ${\mathbb{F}_{q^{2}}}$-maximal curves are given by the Galois ${\mathbb{F}_{q^{2}}}$-subcovers of an ${\mathbb{F}_{q^{2}}}$-maximal curve $\mathcal{X}$, that is by the quotient curves $\mathcal{X}/G$ over a finite automorphism group $G\leq{\rm Aut}(\mathcal{X})$.

A challenging open problem is the determination of the spectrum $\Gamma(q^{2})$ of genera of ${\mathbb{F}_{q^{2}}}$-maximal curves, for a given $q$. Most of the known values in $\Gamma(q^{2})$ have been obtained from quotient curves ${\mathcal{H}}_{q}/G$ of the Hermitian curve, which have been investigated in many papers; the most significant cases are the following:

• •

$G$ fixes an ${\mathbb{F}_{q^{2}}}$-rational point of ${\mathcal{H}}_{q}$; see [2, 13, 4].

• •

$G$ normalizes a Singer subgroup of ${\mathcal{H}}_{q}$ acting on three $\mathbb{F}_{q^{6}}$-rational points of ${\mathcal{H}}_{q}$; see [7, 13].

• •

$G$ has prime order; see [8].

• •

$G$ fixes an ${\mathbb{F}_{q^{2}}}$-rational point off ${\mathcal{H}}_{q}$ and is isomorphic to a subgroup of $SL(2,q)$; see [8].

Let $\mathcal{M}_{q}$ be the maximal subgroup of $PGU(3,q)$ fixing an ${\mathbb{F}_{q^{2}}}$-rational point $P\notin{\mathcal{H}}_{q}$; equivalently, ${\mathcal{M}}_{q}$ fixes the polar line $\ell$ of $P$ with respect to the unitary polarity associated with ${\mathcal{H}}_{q}$. The group-theoretical structure of ${\mathcal{M}}_{q}$ is not known, and only few genera of quotients ${\mathcal{H}}_{q}/G$ with $G\leq{\mathcal{M}}_{q}$ have been computed; see [8, 13]. In this paper, many genera of quotients ${\mathcal{H}}_{q}/G$ with $G\leq{\mathcal{M}}_{q}$ are computed, giving a partial answer to the question raised by Garcia, Stichtenoth, and Xing in [13, Remark 6.6]. To this aim, we provide a new geometric and group-theoretical description of ${\mathcal{M}}_{q}$. This allows us to use the techniques developed in [22] for the determination of the genera of quotients ${\mathcal{H}}_{q}/G$. Some further genera of quotients ${\mathcal{H}}_{q}/G$ with $G\not\leq{\mathcal{M}_{q}}$ are also computed. Our results provide new values in the spectra $\Gamma(q^{2})$. Finally, we consider the quotients ${\mathcal{H}}_{q}/G$ where $G$ is cyclic, and provide a plane model for ${\mathcal{H}}_{q}/G$ in one of the few cases for which an equation is not known, namely when $G$ has order $p\cdot d$ with $d$ a divisor of $q+1$.

2. Preliminary results

Throughout this paper, $q=p^{n}$, where $p$ is a prime number and $n$ is a positive integer. The Deligne-Lusztig curves defined over a finite field $\mathbb{F}_{q}$ were originally introduced in [11]. Other than the projective line, there are three families of Deligne-Lusztig curves, named Hermitian curves, Suzuki curves and Ree curves. The Hermitian curve $\mathcal{H}_{q}$ arises from the algebraic group ${}^{2}A_{2}(q)={\rm PGU}(3,q)$ of order $(q^{3}+1)q^{3}(q^{2}-1)$. It has genus $q(q-1)/2$ and is $\mathbb{F}_{q^{2}}$-maximal. Thus curve is isomorphic to the curves listed below:

[TABLE]

[TABLE]

[TABLE]

where $\omega$ is a fixed element of $\mathbb{K}$ such that $\omega^{q-1}=-1$;

[TABLE]

Each of the models (2.1),(2.2) and (2.3) is $\mathbb{K}$-isomorphic to ${\mathcal{H}}_{q}$, while the model (2.4) is $\mathbb{F}_{q^{3}}$-isomorphic to ${\mathcal{H}}_{q}$, since for a suitable element $a\in\mathbb{F}_{q^{6}}$, the projective map

[TABLE]

changes (2.1) into (2.4), see [7][Proposition 4.6]. The automorphism group ${\rm Aut}({\mathcal{H}}_{q})$ is isomorphic to the projective unitary group ${\rm PGU}(3,q)$, and it acts on the set ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$ of all $\mathbb{F}_{q^{2}}$-rational points of ${\mathcal{H}}_{q}$ as ${\rm PGU}(3,q)$ in its usual $2$-transitive permutation representation. The combinatorial properties of ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$ can be found in [19]. The size of ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$ is equal to $q^{3}+1$, and a line of $PG(2,q^{2})$ has either $1$ or $q+1$ common points with ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$, that is, it is either a $1$-secant or a chord of ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$. Furthermore, a unitary polarity is associated with ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$ whose isotropic points are those of ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$ and isotropic lines are the $1$-secants of ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$, that is, the tangents to ${\mathcal{H}}_{q}$ at the points of ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$.

From Group theory we need the classification of all maximal subgroups of the projective special subgroup ${\rm PSU}(3,q)$ of ${\rm PGU}(3,q)$, going back to Mitchell and Hartley; see [21], [16], [18]; and also the classification of all subgroups of $SL(2,q)$, see [24, Theorem 6.17].

Theorem 2.1**.**

Let $d={\rm gcd}(3,q+1)$. Up to conjugacy, the subgroups below give a complete list of maximal subgroups of ${\rm PSU}(3,q)$.

• (i)

the stabilizer of an ${\mathbb{F}}_{q^{2}}$-rational point of ${\mathcal{H}}_{q}$. It has order $q^{3}(q^{2}-1)/d$;

• (ii)

*the stabilizer of an ${\mathbb{F}}_{q^{2}}$-rational point off ${\mathcal{H}}_{q}$ *$($equivalently the stabilizer of a chord of ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}}))$. It has order $q(q-1)(q+1)^{2}/d$;

• (iii)

the stabilizer of a self-polar triangle with respect to the unitary polarity associated to ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$. It has order $6(q+1)^{2}/d$;

• (iv)

the normalizer of a (cyclic) Singer subgroup. It has order $3(q^{2}-q+1)/d$ and preserves a triangle in ${\rm PG}(2,q^{6})\setminus{\rm PG}(2,q^{2})$ left invariant by the Frobenius collineation $\Phi_{q^{2}}:(X,Y,T)\mapsto(X^{q^{2}},Y^{q^{2}},T^{q^{2}})$ of ${\rm PG}(2,\bar{\mathbb{F}}_{q})$;

for $p>2$:**

• (v)

${\rm PGL}(2,q)$* preserving a conic;*

• (vi)

${\rm PSU}(3,p^{m})$* with $m\mid n$ and $n/m$ odd;*

• (vii)

subgroups containing ${\rm PSU}(3,p^{m})$ as a normal subgroup of index $3$, when $m\mid n$, $n/m$ is odd, and $3$ divides both $n/m$ and $q+1$;

• (viii)

the Hessian groups of order $216$ when $9\mid(q+1)$, and of order $72$ and $36$ when $3\mid(q+1)$;

• (ix)

${\rm PSL(2,7)}$* when $p=7$ or $-7$ is not a square in $\mathbb{F}_{q}$;*

• (x)

the alternating group $Alt(6)$ when either $p=3$ and $n$ is even, or $5$ is a square in $\mathbb{F}_{q}$ but $\mathbb{F}_{q}$ contains no cube root of unity;

• (xi)

the symmetric group $Sym(6)$ when $p=5$ and $n$ is odd;

• (xii)

the alternating group $Alt(7)$ when $p=5$ and $n$ is odd;

for $p=2$:**

• (xiii)

${\rm PSU}(3,2^{m})$* with $m\mid n$ and $n/m$ an odd prime;*

• (xiv)

subgroups containing ${\rm PSU}(3,2^{m})$ as a normal subgroup of index $3$, when $n=3m$ with $m$ odd;

• (xv)

a group of order $36$ when $n=1$.

Theorem 2.2**.**

*Let $q=p^{h}$, where $p$ is a prime and $h\in\mathbb{N}$. Up to isomorphism, the subgroups below give a complete list of subgroups of $SL(2,q)$.

$\bullet$ Tame subgroups:*

1. (1)

$SL(2,5)$, when $p\geq 7$ and $q\equiv 1\ mod\ 5$; 2. (2)

$\hat{\Sigma}_{4}$, the representation group of $S_{4}$ in which the transpositions corerspond to the elements of order $4$, for $p\geq 5$ and $q^{2}\ \equiv 1\ mod\ 16$; this group is isomorphic to $SmallGroup(48,28)$. 3. (3)

$SL(2,3)$, when $p\geq 5$; 4. (4)

$C_{d}$, where $e\mid q\pm 1$; 5. (5)

A dicyclic group $Dic(n)=\langle a,x\mid a^{2n}=1,\ x^{2}=a^{n},\ x^{-1}ax=a^{-1}\rangle$ of order $4n$, where $n\mid\frac{q\pm 1}{2}$;

$\bullet$* Non-tame subgroups:*

1. (1)

$E_{p^{k}}\rtimes C_{d}$, where $d\mid q-1$, and $E_{p^{k}}$ is elementary abelian of order $p^{k}$; 2. (2)

$SL(2,5)$, when $q=3^{h}$, and $5\mid q^{2}-1$; 3. (3)

$SL(2,p^{k})$, where $k\mid h$; 4. (4)

$TL(2,p^{k})\cong\langle SL(2,p^{k}),d_{\pi}\rangle$* where,*

[TABLE]

for $w=\xi^{\frac{p^{k}+1}{2}}$ and $\mathbb{F}^{*}_{p^{2k}}=\langle\xi\rangle$.

In our investigation it is useful to know how an element of ${\rm PGU}(3,q)$ of a given order acts on ${\rm PG}(2,\bar{\mathbb{F}}_{q})$, and in particular on ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$. This can be obtained as a corollary of Theorem 2.1, and is stated in Lemma $2.2$ with the usual terminology of collineations of projective planes; see [19]. In particular, a linear collineation $\sigma$ of ${\rm PG}(2,\bar{\mathbb{F}}_{q})$ is a $(P,\ell)$-perspectivity, if $\sigma$ preserves each line through the point $P$ (the center of $\sigma$), and fixes each point on the line $\ell$ (the axis of $\sigma$). A $(P,\ell)$-perspectivity is either an elation or a homology according as $P\in\ell$ or $P\notin\ell$. A $(P,\ell)$-perspectivity is in ${\rm PGL}(3,q^{2})$ if and only if its center and its axis are in ${\rm PG}(2,\mathbb{F}_{q^{2}})$. This classification results was obtained in [22].

Lemma 2.3**.**

For a nontrivial element $\sigma\in{\rm PGU}(3,q)$, one of the following cases holds.

• (A)

${\rm ord}(\sigma)\mid(q+1)$. Moreover, $\sigma$ is a homology whose center $P$ is a point off ${\mathcal{H}}_{q}$ and whose axis $\ell$ is a chord of ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$ such that $(P,\ell)$ is a pole-polar pair with respect to the unitary polarity associated to ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$.

• (B)

${\rm ord}(\sigma)$* is coprime with $p$. Moreover, $\sigma$ fixes the vertices $P_{1},P_{2},P_{3}$ of a non-degenerate triangle $T$.*

• (B1)

The points $P_{1},P_{2},P_{3}$ are ${\mathbb{F}_{q^{2}}}$-rational, $P_{1},P_{2},P_{3}\notin{\mathcal{H}}_{q}$ and the triangle $T$ is self-polar with respect to the unitary polarity associated to ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$. Also, ${\rm ord}(\sigma)\mid(q+1)$.

• (B2)

The points $P_{1},P_{2},P_{3}$ are ${\mathbb{F}_{q^{2}}}$-rational, $P_{1}\notin{\mathcal{H}}_{q}$, $P_{2},P_{3}\in{\mathcal{H}}_{q}$. Also, ${\rm ord}(\sigma)\mid(q^{2}-1)$ and ${\rm ord}(\sigma)\nmid(q+1)$.

• (B3)

The points $P_{1},P_{2},P_{3}$ have coordinates in ${\mathbb{F}}_{q^{6}}\setminus{\mathbb{F}}_{q^{2}}$, $P_{1},P_{2},P_{3}\in{\mathcal{H}}_{q}$. Also, ${\rm ord}(\sigma)\mid(q^{2}-q+1)$.

• (C)

${\rm ord}(\sigma)=p$. Moreover, $\sigma$ is an elation whose center $P$ is a point of ${\mathcal{H}}_{q}$ and whose axis $\ell$ is a tangent of ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$ such that $(P,\ell)$ is a pole-polar pair with respect to the unitary polarity associated to ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$.

• (D)

${\rm ord}(\sigma)=p$* with $p\neq 2$, or ${\rm ord}(\sigma)=4$ and $p=2$. Moreover, $\sigma$ fixes an ${\mathbb{F}_{q^{2}}}$-rational point $P$, with $P\in{\mathcal{H}}_{q}$, and a line $\ell$ which is a tangent of ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$, such that $(P,\ell)$ is a pole-polar pair with respect to the unitary polarity associated to ${\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$.*

• (E)

$p\mid{\rm ord}(\sigma)$, $p^{2}\nmid{\rm ord}(\sigma)$, and ${\rm ord}(\sigma)\neq p$. Moreover, $\sigma$ fixes two ${\mathbb{F}_{q^{2}}}$-rational points $P,Q$, with $P\in{\mathcal{H}}_{q}$, $Q\notin{\mathcal{H}}_{q}$.

Throughout the paper, a nontrivial element of ${\rm PGU}(3,q)$ is said to be of type (A), (B), (B1), (B2), (B3), (C), (D), or (E), as given in Lemma 2.3. Moreover, $G$ always stands for a subgroup of ${\rm PGU}(3,q)$.

From Function field theory we need the Riemann-Hurwitz formula; see [23, Theorem 3.4.13]. Every subgroup $G$ of ${\rm PGU}(3,q)$ produces a quotient curve ${\mathcal{H}}_{q}/G$, and the cover ${\mathcal{H}}_{q}\rightarrow{\mathcal{H}}_{q}/G$ is a Galois cover defined over $\mathbb{F}_{q^{2}}$ where the degree of the different divisor $\Delta$ is given by the Riemann-Hurwitz formula, namely $\Delta=(2g({\mathcal{H}}_{q})-2)-|G|(2g({\mathcal{H}}_{q}/G)-2)$. On the other hand, $\Delta=\sum_{\sigma\in G\setminus\{id\}}i(\sigma)$, where $i(\sigma)\geq 0$ is given by the Hilbert’s different formula [23, Thm. 3.8.7], namely

[TABLE]

where $t$ is a local parameter at $P$.

By analyzing the geometric properties of the elements $\sigma\in{\rm PGU}(3,q)$, it turns out that there are only a few possibilities for $i(\sigma)$. This is obtained as a corollary of Lemma 2.3 and stated in the following proposition, see [22].

Theorem 2.4**.**

For a nontrivial element $\sigma\in{\rm PGU}(3,q)$ one of the following cases occurs.

1. (1)

If ${\rm ord}(\sigma)=2$ and $2\mid(q+1)$, then $\sigma$ is of type (A) and $i(\sigma)=q+1$. 2. (2)

If ${\rm ord}(\sigma)=3$, $3\mid(q+1)$ and $\sigma$ is of type (B3), then $i(\sigma)=3$. 3. (3)

If ${\rm ord}(\sigma)\neq 2$, ${\rm ord}(\sigma)\mid(q+1)$ and $\sigma$ is of type (A), then $i(\sigma)=q+1$. 4. (4)

If ${\rm ord}(\sigma)\neq 2$, ${\rm ord}(\sigma)\mid(q+1)$ and $\sigma$ is of type (B1), then $i(\sigma)=0$. 5. (5)

If ${\rm ord}(\sigma)\mid(q^{2}-1)$ and ${\rm ord}(\sigma)\nmid(q+1)$, then $\sigma$ is of type (B2) and $i(\sigma)=2$. 6. (6)

If ${\rm ord}(\sigma)\neq 3$ and ${\rm ord}(\sigma)\mid(q^{2}-q+1)$, then $\sigma$ is of type (B3) and $i(\sigma)=3$. 7. (7)

If $p=2$ and ${\rm ord}(\sigma)=4$, then $\sigma$ is of type (D) and $i(\sigma)=2$. 8. (8)

If ${\rm ord}(\sigma)=p$, $p\neq 2$ and $\sigma$ is of type (D), then $i(\sigma)=2$. 9. (9)

If ${\rm ord}(\sigma)=p$ and $\sigma$ is of type (C), then $i(\sigma)=q+2$. 10. (10)

If ${\rm ord}(\sigma)\neq p$, $p\mid{\rm ord}(\sigma)$ and ${\rm ord}(\sigma)\neq 4$, then $\sigma$ is of type (E) and $i(\sigma)=1$.

3. The maximal subgroup ${\mathcal{M}_{q}}$ of ${\rm PGU}(3,q)$ for $q$ odd

The aim of this section is to give an explicit description of the maximal subgroup ${\mathcal{M}_{q}}$ of ${\rm PGU}(3,q)$. Geometrically, ${\mathcal{M}_{q}}$ preserves a non-tangent point-line pair $(P_{0},\ell)$, where $P_{0}\in\mathbb{P}^{2}({\mathbb{F}_{q^{2}}})\setminus{\mathcal{H}}_{q}$ and $\ell$ is its polar line with respect to the unitary polarity associated to ${\mathcal{H}}_{q}$. The line $\ell$ is a $\mathbb{F}_{q^{2}}$-rational line meeting ${\mathcal{H}}_{q}$ in $q+1$ pairwise distinct $\mathbb{F}_{q^{2}}$-rational points. We use the plane model (2.3) of ${\mathcal{H}}_{q}$. Up to conjugation we can choose $P_{0}=(0:0:1)$ and $\ell:Z=0$. Then the subgroup of automorphisms of $\mathbb{P}^{2}(\mathbb{K})$ preserving both $(P_{0},\ell)$ and ${\mathcal{H}}_{q}$, consists of maps of type

[TABLE]

where

[TABLE]

Clearly, $|{\mathcal{M}_{q}}|=q(q-1)(q+1)$. Those maps with $a,b,c,d\in\mathbb{F}_{q}$ and $ad-bc=1$, form a subgroup $H$ of ${\mathcal{M}_{q}}$ which is isomorphic to $SL(2,{\mathbb{F}_{q}})$. By direct checking $H$ is the commutator subgroup of ${\mathcal{M}_{q}}$ and hence it is normal in ${\mathcal{M}_{q}}$. We recall that since $\mathbb{K}$ has odd characteristic, every involution $\tau$ is an homology, see Lemma $2.2$. We can give a geometrical interpretation of ${\mathcal{M}_{q}}$ as ${\mathcal{M}_{q}}=C_{{\rm PGU}(3,q)}(\tau)=N_{{\rm PGU}(3,q)}(\tau)$, where $\tau$ is the $(P_{0},\ell)$-involution given by

[TABLE]

The center of ${\mathcal{M}_{q}}$, say $Z({\mathcal{M}_{q}})$, is a cyclic group of order $q+1$. This subgroup consists of all the $(P_{0},\ell)$-homologies contained in ${\mathcal{M}_{q}}$. Summarizing,

[TABLE]

[TABLE]

[TABLE]

Proposition 3.1**.**

Let $\tau^{\prime}$ be a fixed $(P^{\prime}_{0},\ell^{\prime})$-involution, where $P^{\prime}_{0}\in\ell$, $P^{\prime}_{0}\not\in{\mathcal{H}}_{q}$ and $\ell^{\prime}$ is the polar line of $P^{\prime}_{0}$ with respect to the polarity associated to ${\mathcal{H}}_{q}$. Then

[TABLE]

where $Z(C_{{\rm PGU}(3,q)}(\tau^{\prime}))$ is the center of the centralizer of $\tau^{\prime}$ in ${\rm PGU}(3,q)$.

Proof.

Since $C_{{\rm PGU}(3,q)}(\tau^{\prime})$ is a subgroup of ${\rm PGU}(3,q)$ which is conjugated to ${\mathcal{M}_{q}}$, we know that $Z(C_{{\rm PGU}(3,q)}(\tau^{\prime}))$ is the group of all the $(P^{\prime}_{0},\ell^{\prime})$-homologies and hence it is cyclic of order $q+1$ and since $Z(C_{{\rm PGU}(3,q)}(\tau^{\prime}))$ fixes $P_{0}$, we have that $Z(C_{{\rm PGU}(3,q)}(\tau^{\prime}))\subseteq{\mathcal{M}_{q}}$. Since $H=({\mathcal{M}_{q}})^{\prime}$, the product $HZ(C_{{\rm PGU}(3,q)}(\tau^{\prime}))$ is a subgroup of ${\mathcal{M}_{q}}$. Moreover, $Z(C_{{\rm PGU}(3,q)}(\tau^{\prime}))\cap H=\{1\}$ because, looking at the matrix representations, every element of $H$ has determinant equal to $1$. Thus, $HZ(C_{{\rm PGU}(3,q)}(\tau^{\prime}))=H\rtimes Z(C_{{\rm PGU}(3,q)}(\tau^{\prime}))$ and since $|HZ(C_{{\rm PGU}(3,q)}(\tau^{\prime}))|=|H|(q+1)=|{\mathcal{M}_{q}}|$, the claim follows. ∎

We can give an explicit matrix representation for ${\mathcal{M}_{q}}$ as follows. In Proposition 3.1 we prove that a complement for $H$ in ${\mathcal{M}_{q}}$ is given by a cyclic group of order $q+1$ given by $(P^{\prime}_{0},\ell^{\prime})$-homologies, where $P^{\prime}_{0}\in\ell$, $P^{\prime}_{0}\not\in{\mathcal{H}}_{q}$ and $\ell^{\prime}$ is the polar line of $P^{\prime}_{0}$ with respect to the polarity associated to ${\mathcal{H}}_{q}$. We can construct such a complement fixing a collineation $\sigma\in PGU(3,q)$ such that $\sigma(P_{0})=P^{\prime}_{0}$. Clearly, $Z(C_{{\rm PGU}(3,q)}(\tau))^{\sigma}=Z(C_{{\rm PGU}(3,q)}(\tau^{\prime}))$ and $H\cap Z(C_{{\rm PGU}(3,q)}(\tau))^{\sigma}=\{1\}$. In this way the following corollary is obtained.

Corollary 3.2**.**

Let $e$ be a primitive element of $\mathbb{F}_{q^{2}}$. Then

[TABLE]

where, $\alpha$ is associated to the matrix representation

[TABLE]

The following lemma collects geometric properties of the elements of ${\mathcal{M}_{q}}$. It is obtained as a direct consequence of Lemma $2.2$ and Theorem $2.7$.

Lemma 3.3**.**

Let $\sigma\in{\mathcal{M}_{q}}$. Then,

1. (1)

If $\sigma$ is a $p$-element then $\sigma$ is of type $(C)$ and $i(\sigma)=q+2$; 2. (2)

If $\sigma\in H$ and $2\neq o(\sigma)\mid q+1$ then $\sigma$ is of type $(B1)$ and $i(\sigma)=0$.

Proof.

Let $\sigma\in{\mathcal{M}_{q}}$ be a $p$-element. From Lemma $2.2$ $\sigma$ is either of type $(C)$ or $(D)$. Since $\sigma$, by definition of ${\mathcal{M}_{q}}$, must fix $P_{0}\not\in{\mathcal{H}}_{q}$ then $\sigma$ cannot be of type $(D)$. Now, $(1)$ follows from Theorem $2.7$. Let $\sigma\in{\mathcal{M}_{q}}$ with $o(\sigma)\neq 2$ but $o(\sigma)\mid q+1$. Since, looking at the matrix representation of $\sigma$, we have that $det(\sigma)=1$, then

[TABLE]

where $o(a)=o(\sigma)$. Thus, by direct checking $\sigma$ is of type $(B1)$, and so $i(\sigma)=0$ from Theorem $2.7$. ∎

Let $G$ be a subgroups of $H$. The following lemma shows that the action of $G$ on the affine points of ${\mathcal{H}}_{q}$ is semi-regular, i.e. each point-orbit of affine points of ${\mathcal{H}}_{q}$ under $G$ has length equal to the order of $G$, see [7].

Lemma 3.4**.**

Let $\alpha\in H$ and $P\in{\mathcal{H}}_{q}$ an affine point such that $\alpha(P)=P$. Then $\alpha$ is the identity map.

Proof.

It follows from $(\ref{map})$ and the fact that $a^{q}=a$ for each $a\in\mathbb{F}_{q}$. ∎

Now we investigate the action of $G$ on the set $\mathcal{I}=\ell\cap{\mathcal{H}}_{q}$ consisting of all points $(1:m:0)$, with $m\in\mathbb{F}_{q}$, together with $(0:1:0)$. Since $H$ acts on $\mathcal{I}$ as $PSL(2,q)$ in its natural $2$-transitive permutation representation on $\mathbb{P}^{1}(\mathbb{F}_{q})$, we have actually to consider $\bar{G}$ instead of $G$, where $\bar{G}$ is the image of $G$ under the canonical epimorphism

[TABLE]

Note that the kernel of $\phi$ is trivial for $p=2$, otherwise it is the subgroup of order $2$ generated by the involution $\tau$. Hence either $ord(G)=2ord(\bar{G})$ or $ord(G)=ord(\bar{G})$, and in the later case $ord(\bar{G})$ must be odd.

The tame subgroups of $H$ are analyzed in [7], using this equivalent representation. In the following section a large class of tame and non-tame subgroups of ${\mathcal{M}_{q}}$ is studied.

4. Genera of quotient curves ${\mathcal{H}}_{q}/H$, for some $H\leq{\mathcal{M}_{q}}$

In this section our aim is to compute the genus of the quotient curve of ${\mathcal{H}}_{q}$ arising from those subgroups of ${\mathcal{M}_{q}}$ that can be written, up to isomorphism, as an internal semidirect product of a subgroup of $H$ and a subgroup of $Z(C_{{\rm PGU}(3,q)}(\tau^{\prime}))\cong C_{q+1}$. To this purpose, we use the same notation and representation for ${\mathcal{M}_{q}}$ introduced in Section 3.

4.1. Non-tame subgroups of ${\bf H\cong SL(2,q)}$

This section provides the genus of ${\mathcal{H}}_{q}/G$ for all non-tame subgroups of $H$, using the complete classification of subgroups of $SL(2,q)$ given in Theorem 2.2. When $G\leq H$ is tame, the genus of ${\mathcal{H}}_{q}/G$ was already obtained in [8, Section 3].

Since ${\mathcal{H}}_{q}$ has zero $p$-rank, each $p$-subgroup has a unique fixed point on ${\mathcal{H}}_{q}$, see [17, Lemma 11.129]. Let $G$ be the subgroup of $H\cong SL(2,q)$ given by $E_{p^{k}}\rtimes C_{d}$, where $d\mid q-1$. Since $E_{p^{k}}$ is a normal subgroup of $G$, then $G$ fixes a point $P\in{\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$. In this case the computation of the genus of the quotient curve ${\mathcal{H}}_{q}/G$ is computed in [13].

Proposition 4.1**.**

([13, Theorem 4.4])* Let $G\leq H$ with $G\cong E_{p^{k}}\rtimes C_{d}$ with $d\mid(q-1)$. Then the genus $\bar{g}$ of the quotient curve ${\mathcal{H}}_{q}/G$ is*

[TABLE]

Proposition 4.2**.**

Let $G\leq H$ with $G\cong SL(2,5)$, for $q=3^{h}$ and $5\mid q^{2}-1$. Then the genus $\bar{g}$ of the quotient curve ${\mathcal{H}}_{q}/G$ is given by:

[TABLE]

Proof.

By direct computation we get that $5\mid q^{2}-1$ if and only if $q=3^{h}$ where $h$ is even. We can apply Theorem 2.4 and Lemma 3.3 looking at the congruence of $h$ modulo $4$; in this way the degree of the different divisor is now obtained simply analyzing the order statistic of the elements of $G$. Assume that $h\equiv 0\ mod\ 4$. By direct checking $5\mid q-1$ and so also $10\mid q-1$. From Theorem 2.4 we have

[TABLE]

Using the same argument we get, if $h\equiv 2\ mod\ 4$:

[TABLE]

Since $SL(2,5)$ contains exactly $1$ element of order $2$, $20$ elements of order $3$, $24$ elements of order $5$, $30$ elements of order $4$, $20$ elements of order $6$ and $24$ elements of order $10$, the claim follows as a direct application of the Riemann-Hurwitz formula. ∎

Proposition 4.3**.**

Let $G\leq H$ with $G\cong SL(2,\bar{q})$, for $\bar{q}=p^{k}$, $k\leq h$. Then the genus $\bar{g}$ of the quotient curve ${\mathcal{H}}_{q}/G$ is given by:

[TABLE]

Proof.

Suppose that $q=\bar{q}$. Looking at the structure of the conjugacy classes of $SL(2,q)$ (see for instance [24, §3.6]) we get the following classification for the elements of $SL(2,q)$:

• •

$1$ element $\alpha$ of order $2$. Thus, $i(\alpha)=q+1$, from Theorem 2.4;

• •

$\frac{q(q-1)^{2}}{2}$ nontrivial elements whose order is different from $2$ but divides $q+1$. From Lemma 3.3 the contribution to the degree of the different divisors of these elements is equal to [math];

• •

$\frac{q(q+1)(q-3)}{2}$ nontrivial elements whose order is different from $2$ but divides $q-1$. From Theorem 2.4 the contribution to the degree of the different divisors of these elements is equal to $2$;

• •

$(q-1)(q+1)$ elements of order $p$. From Lemma 3.3 the contribution of these elements is equal to $q+2$;

• •

$q^{2}-1$ elements of order a multiple of $p$ different from $p$. From Theorem 2.4 their contribution to the different divisor is equal to $1$.

Suppose that $q=\bar{q}^{2r+1}$, so that $\bar{q}+1\mid q+1$ and $\bar{q}-1\mid q-1$. From the Riemann-Hurwitz formula and Theorem 2.4 we get that

[TABLE]

[TABLE]

and the claim follows by direct computation.

Suppose that $q=\bar{q}^{2r}$, so that $\bar{q}-1,\bar{q}+1\mid q-1$. From the Riemann-Hurwitz formula and Theorem 2.4 we get that

[TABLE]

[TABLE]

∎

Proposition 4.4**.**

Let $G\cong TL(2,p^{k})=\langle SL(2,p^{k}),d_{\pi}\rangle$ where,

[TABLE]

for $w=\xi^{\frac{p^{k}+1}{2}}$ and $\mathbb{F}^{*}_{p^{2k}}=\langle\xi\rangle$. Then the genus $\bar{g}$ of the quotient curve ${\mathcal{H}}_{q}/G$ is given by:

[TABLE]

Proof.

Let $\bar{q}=p^{k}$. Since $d_{\pi}\in SL(2,\bar{q}^{2})\setminus SL(2,\bar{q})$, $q$ is an even power $\bar{q}^{2r}$ of $\bar{q}$. The values $i(\sigma)$ for $\sigma\in SL(2,\bar{q})$ are computed in Proposition 4.3. Since $o(d_{\pi})=2(\bar{q}-1)$, we have that $o(d_{\pi})\mid(\bar{q}^{2}-1)$ and $o(d_{\pi})\nmid(\bar{q}+1)$. Thus, according to Theorem 2.4 we have that $i(d_{\pi})=2$. We need to compute $i(\alpha{d_{\pi}}^{s})$ for every $\alpha\in SL(2,\bar{q})$ and $s\in\mathbb{N}$. Note that $q$ is an even power $\bar{q}^{2r}$ of $\bar{q}$, otherwise $d_{\pi}\notin H$; hence $\bar{q}^{2}-1$ divides $q-1$. We show that any element $\alpha d_{\pi}$ with $\alpha\in SL(2,\bar{q})$ is of type (B2). Suppose $o(\alpha d_{\pi})=2$; then $\alpha d_{\pi}$ is the central involution $\eta$ of $SL(2,\bar{q})$, a contradiction to $d_{\pi}\notin SL(2,\bar{q})$. Suppose $o(\alpha d_{\pi})=4$. Then $(\alpha d_{\pi})^{2}=\gamma d_{\pi}^{2}$ ($\gamma\in SL(2,\bar{q})$) is equal to $\eta$ and hence acts trivially on the line $\ell$; this implies that $\gamma$ is the unique element $(d_{\pi}^{2})^{-1}\in SL(2,\bar{q})$ acting as the inverse of $d_{\pi}^{2}$ on $\ell$, a contradiction. Hence $o(\alpha d_{\pi})>4$. Since $(\alpha d_{\pi})^{2(\bar{q}-1)}\in SL(2,\bar{q})$ is either trivial or of type (B2), the claim follows. From the Riemann-Hurwitz formula

[TABLE]

[TABLE]

∎

4.2. Some tame subgroups of ${\mathcal{M}_{q}}$

We refer to the complete classification of subgroups of $SL(2,q)$ given in Theorem 2.2. In this Section the genus of every quotient curve ${\mathcal{H}}_{q}/G$, for $G=N\rtimes C$ with $N<H$ and $C\cong C_{m}$, $m\mid(q+1)$, is computed.

Proposition 4.5**.**

Let $G=C_{d}\rtimes C_{m}=\langle\beta\rangle\rtimes\langle\alpha\rangle$ where $C_{d}<H$, $d\mid(q-1)$, and $m\mid(q+1)$ of ${\mathcal{M}_{q}}$ then the genus $\bar{g}$ of the quotient curve ${\mathcal{H}}_{q}/G$ is given by one of the following values:

[TABLE]

Proof.

Case 1. d is odd and $\bf m=2$  

From Lemma 2.3 and Theorem 2.4, every $\sigma\in C_{d}$ is of type (B2) and hence $i(\sigma)=2$. Let $P,Q$ and $R$ be the fixed points of $\beta$, with $P,Q\in{\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$ and $R\not\in{\mathcal{H}}_{q}$. Assume that $\alpha$ and $\beta$ commute. Since $d$ and $p$ are odd, $G=C_{2d}$ and $2d\mid(q-1)$. Every element $\gamma=\sigma\alpha$ with $\sigma\in C_{d}$ is of type (B2) as $\gamma$ has exactly the same fixed points of $\sigma$. Proposition 2.4 and the Riemann-Hurwitz formula yield

[TABLE]

and the claim follows. Assume that $\alpha$ and $\beta$ do not commute. Thus, $\alpha(P)=Q$ and $\alpha(R)=R$. Let $\sigma\in C_{d}$ with $o(\sigma)=d^{\prime}\mid d$. Up to conjugation we can assume $P=(0:0:1)$ and $Q=(1:0:0)$ where ${\mathcal{H}}_{q}$ has equation (2.2). Equivalently up to conjugation, $\sigma$ is given by the following matrix representation,

[TABLE]

for some $a\in\mathbb{F}_{q^{2}}$ with $o(a)=d^{\prime}$, and hence

[TABLE]

By direct computation $o(\sigma\alpha)=2\frac{o(\sigma)}{(o(\sigma),q-1)}$ and since $d^{\prime}\mid(q-1)$ we have $o(\sigma\alpha)=2$ for every $\sigma\in C_{d}$. This proves that $G$ is a dihedral group of order $2d$, and Theorem 2.4 and the Riemann-Hurwitz formula yield

[TABLE]

and the claim follows.  

Case 2. d is even and $\bf m=2$  

From Lemma 2.3, $C_{d}$ contains a unique element of order $2$ which is a $(R,\ell_{R})$-homology. Thus, $\beta$ and $\alpha$ cannot commute since otherwise $C_{d}$ and $C_{m}$ are not disjoint. Arguing as in Case 1, $G$ is a dihedral group of order $2d$ and hence from the Riemann-Hurwitz formula

[TABLE]

Case 3. d is odd and $\bf m>2$  

Assume that $\alpha$ and $\beta$ commute; equivalently, assume that $\alpha$ is a $(R,\ell_{R})$-homology. From Theorem 2.3 and the Riemann-Hurwitz formula

[TABLE]

Thus, assume that $\beta$ and $\alpha$ do not commute. Hence $\alpha(P)=Q$ and $\alpha(R)=R$. Since $C_{m}$ has an orbit of length $2$, $m$ must be even and $\alpha^{2}$ is a $(R,\ell_{R})$-homology of order $m/2$. We note that in this case $\alpha$ and the other elements of $C_{m}$ which lie outside $\langle\alpha^{2}\rangle$ must be of type (B1) since $m>2$ and a homology acts semiregularly outside its center and axis. Let $\alpha^{\prime}\in C_{m}\setminus\langle\alpha^{2}\rangle$ and $\beta^{\prime}\in C_{d}$. As before we can assume up to conjugation that $\beta^{\prime}$ is given by the following matrix representation,

[TABLE]

for some $a\in\mathbb{F}_{q^{2}}$ with $o(a)=o(\beta^{\prime})$, and hence

[TABLE]

for some $A,B\in\mathbb{F}_{q^{2}}^{*}$. Thus,

[TABLE]

This prove that $o(\beta^{\prime}\alpha^{\prime})=2o(AB)=o(\alpha^{\prime})\mid(q+1)$ and thus $\beta^{\prime}\alpha^{\prime}$ is either of type (B1) or of type (A). In particular from the matrix representation of $\beta^{\prime}\alpha^{\prime}$ we get that $\beta^{\prime}\alpha^{\prime}$ is of type (A) if and only if $o(\beta^{\prime}\alpha^{\prime})=2$ and so if and only if $o(\alpha^{\prime})=2$. Now the Riemann-Hurwitz formula yields

[TABLE]

where $\delta$ satisfies

[TABLE]

Case 4. d is even and $\bf m>2$  

In this case $C_{d}$ has exactly one element of order $2$ which is of type (A), and $d-2$ nontrivial elements of type (B2). Assume that $\alpha$ and $\beta$ commute. Since $C_{d}$ and $C_{m}$ are disjoint this case can happen if and only if $m$ is odd. Arguing as before, from the Riemann-Hurwitz formula

[TABLE]

Arguing as in Case 3, $m/2$ must be odd and the Riemann-Hurwitz formula yields

[TABLE]

[TABLE]

∎

Remark 4.6*.*

From the proof of Proposition 4.5 follows that the values given for $\bar{g}$ in Proposition 4.5 are exactly all the possible values of $g({\mathcal{H}}_{q}/G)$ for $G\leq{\mathcal{M}_{q}}$, $G\cong C_{d}\rtimes C_{m}$, $d\mid(q-1)$, $m\mid(q+1)$. In fact, a matrix representation for the generators of $G$ can be explicitly provided as follows, where ${\mathcal{H}}_{q}$ has equation (2.2):

• •

if $\alpha\beta=\beta\alpha$, then $\beta={\rm diag}[a^{q+1},a,1]$ and $\alpha={\rm diag}[1,A,1]$;

• •

if $\alpha\beta\neq\beta\alpha$, then $\beta={\rm diag}[a^{q+1},a,1]$ and

[TABLE]

Proposition 4.7**.**

Let $G=C_{d}\rtimes C_{m}=\langle\beta\rangle\rtimes\langle\alpha\rangle$ where $C_{d}<H$, $d\mid(q+1)$, $d\neq 2$, $m\mid(q+1)$ of ${\mathcal{M}_{q}}$ such that $\alpha$ is of type (A); then the genus $\bar{g}$ of the quotient curve ${\mathcal{H}}_{q}/G$ is given by one of the following values:

[TABLE]

where $e=\sum_{d^{\prime}\mid(m,d)}\varphi(d^{\prime})$ and $\varphi(d^{\prime})$ denotes the Euler Totient function of $d^{\prime}$.

Proof.

Case 1. d is odd and $\bf m=2$  

Since $C_{d}<H$ and $d$ is odd, every $\sigma\in C_{d}$ is of type (B1) from Lemma 3.3. Let $\{P,Q,R\}$ be the fixed points of $\beta$. Assume that $\alpha$ and $\beta$ commute. This is equivalent to require that $\alpha$ is a homology of center $T$ where $T\in\{P,Q,R\}$. Since $d$ and $m$ are coprime, every element of type $\beta^{\prime}\alpha$ is of type (B1) because it fixed $P,Q$ and $R$ but it cannot fixed other points. From the Riemann-Hurwitz formula and Theorem 2.4 we have

[TABLE]

Assume that $\alpha$ and $\beta$ do not commute. In this case we can assume that $\alpha(P)=Q$ and $\alpha(R)=R$. Up to conjugation we can assume that $P=(1:0:0)$, $Q=(0:1:0)$, $R=(0:0:1)$, where ${\mathcal{H}}_{q}$ has equation (2.1). Thus, if $\beta^{\prime}\in C_{d}$ then $\beta^{\prime}$ admits a matrix representation $[\lambda,\lambda^{-1},1]$ for some $\lambda\in\mathbb{F}_{q^{2}}$ with $o(\lambda)=o(\beta^{\prime})$, and hence

[TABLE]

By direct checking $o(\beta^{\prime}\alpha)=2$ and $G$ is a dihedral group. Now, from the Riemann-Hurwitz formula

[TABLE]

Case 2. d is odd and $\bf m>2$  

Assume that $\alpha$ and $\beta$ commute. This is equivalent to require that $\alpha$ is a homology of center $T$ where $T\in\{P,Q,R\}$. Let $D=(d,m)$. As before we can assume up to conjugation that $P=(1:0:0)$, $Q=(0:1:0)$, $R=(0:0:1)$ and ${\mathcal{H}}_{q}$ has equation (2.1). Thus $\beta=[\lambda,\lambda^{-1},1]$ and $\alpha$ is either $\alpha=[\gamma,\gamma,1]$, or $\alpha=[1,\gamma,1]$ or $\alpha=[1,1,\gamma]$; say $\alpha=[1,\gamma,1]$. By direct checking, $\beta^{h}\alpha^{k}$ is of type (A) if and only if $\gamma^{k}\in\{\lambda^{h},\lambda^{2h}\}$, otherwise $\beta^{h}\alpha^{k}$ is of type (B1). This proves that if $D\neq 1$ then for every $d^{\prime}\mid D$, $G$ contains $2\varphi(d^{\prime})$ elements of type (A) other than the elements of $C_{m}$; as usual, set $\varphi(1)=0$. From the Riemann-Hurwitz formula

[TABLE]

We note that in Case 2 $\alpha$ and $\beta$ must commute, since $\alpha$ acts semiregularly outside its center and axis while $m>2$.  

Case 3. d is even and $\bf m=2$  

In this case $C_{d}$ contains exactly one element of type (A) which has order $2$, and the other $d-2$ elements are of type (B1). Assume that $\alpha$ and $\beta$ commute. Arguing with the matrix representations as in Case 2, $G$ contains just a homology other than $\beta^{o(\beta)/2}$ and $\alpha$ which is given by $\beta^{o(\beta)/2}\alpha$ and the other elements are of type (B1). By the Riemann-Hurwitz formula

[TABLE]

Assume that $\alpha$ and $\beta$ do not commute. Using the matrix representation of Case 1, one can check that also in this case $G$ is a dihedral group. Thus,

[TABLE]

Case 4. d is even and $\bf m>2$  

Since $m>2$ and $\alpha$ acts semiregularly outside its center and axis, $\alpha$ and $\beta$ commute. Arguing as in Case 2, the Riemann-Hurwitz formula yields

[TABLE]

∎

Remark 4.8*.*

As in Remark 4.6, Proposition 4.7 provides the genus of ${\mathcal{H}}_{q}/G$ for any $G$ as in the hypothesis of Proposition 4.7; conversely, a group $G$ such that the genus of ${\mathcal{H}}_{q}/G$ is any of the genera stated in Proposition 4.7 exists.

Proposition 4.9**.**

Let $G=SL(2,5)\rtimes C_{m}\leq{\mathcal{M}_{q}}$ where $p\geq 7$, $q\equiv 1\pmod{5}$, $\langle\alpha\rangle=C_{m}$ and $m\mid(q+1)$ such that $\alpha$ is of type (A); then the genus $\bar{g}$ of the quotient curve ${\mathcal{H}}_{q}/G$ is given by one of the following values

[TABLE]

where $D=(m,3)$.

Proof.

Table 1, as a direct application of Theorem 2.3 summarizes the values of $i(\sigma)$, for $\sigma\in SL(2,5)$ according to the congruence of $q$ modulo $12$.

Assume that $\alpha$ and $SL(2,5)$ commute. Thus, $\alpha\in Z({\mathcal{M}_{q}})$. Since $|SL(2,5)|$ is even, then $m$ must be odd since otherwise $SL(2,5)$ and $C_{m}$ cannot be disjoint. Assume that $q\equiv_{12}1$. From the Riemann-Hurwitz formula

[TABLE]

[TABLE]

now the claim follows by direct computation. Assume that $q\equiv_{12}7$. From the Riemann-Hurwitz formula

[TABLE]

[TABLE]

Assume that $q\equiv_{12}5$. Since $m$ is odd and we can get a homology of the form $\beta\alpha$, for $\beta\in SL(2,5)$ if and only if $o(\beta)=o(\alpha)$, we can construct homologies $\beta\alpha\in G$ if and only if $o(\alpha)=o(\beta)=3$. Denote $D=(m,3)$. From the Riemann-Hurwitz formula

[TABLE]

[TABLE]

Assume that $q\equiv_{12}11$. Arguing as in the previous case, from the Riemann-Hurwitz formula

[TABLE]

[TABLE]

We now assume that $\alpha\not\in Z({\mathcal{M}_{q}})$. Let $q\equiv_{12}1$. From [25, Proposition 1.2], if $m$ is odd then $G$ is a direct product of $SL(2,5)$ and $C_{m}$ and hence $\alpha\in Z({\mathcal{M}_{q}})$, a contradiction. We assume that $m$ is even. Since $2\mid(q+1)$ but $4\nmid(q+1)$, we can write $m=2\tilde{m}$, where $\tilde{m}$ is odd. If $\tilde{m}>1$ then $C_{m}=C_{\tilde{m}}\times C_{2}$ is a cyclic group generated by a homology $\alpha$, which has two proper subgroups $C_{\tilde{m}}$ and $C_{2}$ of homologies with different axes; a contradiction. Hence $m=2$ and $G=SL(2,5)\rtimes C_{2}$. By direct checking, with MAGMA, there are just 4 group of order 240 containing $SL(2,5)$, namely $G\cong SmallGroup(240,i)$ with $i\in\{89,90,93,94\}$.

• •

$i=89$: cannot occur, since in this case $G$ must contain a unique involution; a contradiction.

• •

$i=93$: then $G$ contains a cyclic normal subgroup $C_{4}=\langle\beta\rangle$ of order $4$. From Theorem 2.3, $\beta$ is of type (B2) fixing three $\mathbb{F}_{q^{2}}$-rational points $P,Q,R$ such that $P,Q\in{\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$ and $R\not\in{\mathcal{H}}_{q}$. Every $\sigma\in G$ normalizes $\beta$ and hence either fixes $P$ and $Q$ or $\sigma(P)=Q$. Since every element of order $3$ must commute with $\beta$, because it cannot have an orbit of length $2$, we have a contradiction.

• •

$i=94$: then $G=SL(2,5)\times C_{2}$; a contradiction.

• •

$i=90$: from the Riemann-Hurwitz formula,

[TABLE]

since by direct computation $\bar{g}\not\in\mathbb{Z}$, this case is impossible.

Assume that $q\equiv_{12}7$. Arguing as in the previous case $G=SL(2,5)\rtimes C_{2^{h}}$ where $h\geq 1$ and $2^{h}\mid(q+1)$. Then $\alpha$ acts on a set $I$ of $\mathbb{F}_{q^{2}}$-rational triangles which are the fixed points of the subgroups of order $4$ of $SL(2,5)$, which have not common fixed points as they do not commute. Since $|I|=15$ is odd, $\alpha$ must have at least a fixed triangle $T=\{P,Q,R\}$. Since $\alpha\not\in Z({\mathcal{M}_{q}})$ then $\alpha(P)=Q$ and hence there is a $\alpha$-orbit of length $2$. This implies that $o(\alpha)=2$ and $G=SL(2,5)\rtimes C_{2}$. As before $G\cong SmallGroup(240,i)$ for $i\in\{89,90,93,94\}$; arguing as above, we get a contradiction in each case. The proofs for $q\equiv_{12}5$ and $q\equiv_{12}11$ are similar to the previous ones. ∎

Remark 4.10*.*

As in Remark 4.6, Proposition 4.9 provides the genus of ${\mathcal{H}}_{q}/G$ for any $G$ as in the hypothesis of Proposition 4.9; conversely, a group $G$ such that the genus of ${\mathcal{H}}_{q}/G$ is any of the genera stated in Proposition 4.7 exists.

Proposition 4.11**.**

Let $G=SL(2,3)\rtimes C_{m}$ of ${\mathcal{M}_{q}}$ where $p\geq 5$, $\langle\alpha\rangle=C_{m}$ and $m\mid(q+1)$ such that $\alpha$ is of type (A); then the genus $\bar{g}$ of the quotient curve ${\mathcal{H}}_{q}/G$ is given by one of the following values:

[TABLE]

where $D=(m,3)$ and $S=0$ if $8\mid(q+1)$, while $S=2$ if $8\nmid(q+1)$.

Proof.

Table 2, as a direct application of Theorem 2.3 summarizes the values of $i(\sigma)$, for $\sigma\in SL(2,3)$ according to the congruence of $q$ modulo $12$.

Assume that $\alpha$ and $SL(2,3)$ commute. Thus, $\alpha\in Z({\mathcal{M}_{q}})$. Since $|SL(2,3)|$ is even, then $m$ must be odd since otherwise $SL(2,3)$ and $C_{m}$ cannot be disjoint. Assume that $q\equiv_{12}1$. From the Riemann-Hurwitz formula

[TABLE]

[TABLE]

now the claim follows by direct computation. Assume that $q\equiv_{12}7$. From the Riemann-Hurwitz formula

[TABLE]

[TABLE]

Assume that $q\equiv_{12}5$. Since $m$ is odd and we can get a homology of the form $\beta\alpha$, for $\beta\in SL(2,5)$ if and only if $o(\beta)=o(\alpha)$, we can construct homologies $\beta\alpha\in G$ if and only if $o(\alpha)=o(\beta)=3$. Denote $D=(m,3)$. From the Riemann-Hurwitz formula

[TABLE]

[TABLE]

Assume that $q\equiv_{12}11$. Arguing as in the previous case, from the Riemann-Hurwitz formula

[TABLE]

[TABLE]

We now assume that $\alpha\not\in Z({\mathcal{M}_{q}})$. From [25, Proposition 1.2], if $m$ is odd then $G$ is a direct product of $SL(2,3)$ and $C_{m}$ and hence $\alpha\in Z({\mathcal{M}_{q}})$, a contradiction. We assume that $m$ is even. If either $q\equiv_{12}1$ or $q\equiv_{12}5$ then since $2\mid(q+1)$ but $4\nmid(q+1)$, we can write $m=2\tilde{m}$, where $\tilde{m}$ is odd. If $\tilde{m}>1$ then $C_{m}=C_{\tilde{m}}\times C_{2}$ is a cyclic group generated by a homology $\alpha$, which has two proper subgroups $C_{\tilde{m}}$ and $C_{2}$ of homologies with different axis; a contradiction. Hence $m=2$ and $G=SL(2,3)\rtimes C_{2}$. By direct checking, with MAGMA, there are just 4 group of order 48 containing $SL(2,3)$, namely $G\cong SmallGroup(48,i)$ for some $i\in\{28,29,32,33\}$.

• •

$i=28$: this case cannot occur since $G$ has more than $1$ involution.

• •

$i=29$: from the Riemann-Hurwitz formula

[TABLE]

if $q\equiv_{12}1$, and

[TABLE]

if $q\equiv_{12}5$.

• •

$i=32$: this case cannot occur as $G\not\cong SL(2,3)\times C_{2}$.

• •

$i=33$: then $G$ contains a cyclic normal subgroup $C_{4}=\langle\beta\rangle$ of order $4$. From Theorem 2.3, $\beta$ is of type (B2) fixing three $\mathbb{F}_{q^{2}}$-rational points $P,Q,R$ such that $P,Q\in{\mathcal{H}}_{q}(\mathbb{F}_{q^{2}})$ and $R\not\in{\mathcal{H}}_{q}$. Every $\sigma\in G$ normalizes $\beta$ and hence either fixes $P$ and $Q$ or $\sigma(P)=Q$. Since every element of order $3$ must commute with $\beta$, because it cannot have an orbit of length $2$, we have a contradiction.

Assume that either $q\equiv_{12}7$ or $q\equiv_{12}11$. Arguing as in the previous case $G=SL(2,3)\rtimes C_{2^{h}}$ where $h\geq 1$ and $2^{h}\mid(q+1)$. Assume that there exists $\alpha^{\prime}\in C_{2^{h}}$ such that $4\mid o(\alpha^{\prime})$. Then $\alpha^{\prime}$ acts on a set of $6$ triangles of type (B1), which are the fixed points of the elements of order $4$ in $SL(2,3)$, without fixed triangles. Since this set is made of $12$ points and a homology acts semiregularly outside its center and axis, we get that $o(\alpha^{\prime})=4$. Thus either $G=SL(2,3)\rtimes C_{2}$ or $G=SL(2,3)\rtimes C_{4}$. We consider the case in which $G\cong SL(2,3)\rtimes C_{2}$. By direct checking, with MAGMA, there are just 4 group of order 48 containing $SL(2,3)$, namely $G\cong SmallGroup(48,i)$ for some $i\in\{28,29,32,33\}$.

• •

$i=28$: this case cannot occur since $G$ has more than $1$ involution.

• •

$i=29$: since there exists $\sigma\in G$ of order $8$ such that $\sigma^{2}$ is of type (B1), then a necessary condition is $8\mid(q+1)$. Let $\sigma\in G$ such that $o(\sigma)=8$. Then $\sigma$ can be either of type (A) or of type (B1). Since $\sigma^{2}$ is of type (B1) then $\sigma$ cannot be of type (A). From the Riemann-Hurwitz formula

[TABLE]

for $q\equiv_{12}7$, and

[TABLE]

for $q\equiv_{12}11$. The claim now follows by direct computation.

• •

$i=32$: this case cannot occur as $G\not\cong SL(2,3)\times C_{2}$.

• •

$i=33$: As before, when $q\equiv_{12}11$, the elements of order $12$ are of type (B1) as they are all conjugated in $G$ and powers of some of them are of type (B1). The group $G$ contains a cyclic group of order $4$ which is not contained in $SL(2,3)$ and it is not conjugated to the other cyclic subgroups of order $4$. A direct computation shows that its generators are of type (A), otherwise from the Riemann-Hurwitz formula $\bar{g}\not\in\mathbb{Z}$. From the Riemann-Hurwitz formula

[TABLE]

if $q\equiv_{12}7$ and

[TABLE]

if $q\equiv_{12}11$. The claim now follows by direct computation.

We consider the case in which $G\cong SL(2,3)\rtimes C_{4}$. By direct checking with MAGMA, there are 4 groups of order 96 containing $SL(2,3)$ as a normal subgroup with $G/SL(2,3)\cong C_{4}$, namely $G\cong SmallGroup(96,i)$ for some $i\in\{66,67,69,74\}$.

• •

$i=66$ or $i=69$: in this case $G$ has three normal subgroups of order $2$. Therefore $G$ fixes pointwise a self-polar triangle and is abelian, a contradiction.

• •

$i=67$: since $Z(G)\cong C_{4}$, looking at the conjugacy classes of $G$, from the Riemann-Hurwitz formula we have

[TABLE]

[TABLE]

if $q\equiv_{12}7$, and

[TABLE]

[TABLE]

if $q\equiv_{12}11$. The claim now follows by direct checking.

• •

$i=74$: since there exists $\sigma\in G$ of order $8$ such that $\sigma^{2}$ is of type (B1), then a necessary condition is $8\mid(q+1)$. Moreover, $G$ contains 16 elements of order $8$: 12 of them are of type (B1) and the remaining 4 are of type (A). In fact $Z(G)\cong C_{8}$, and so $G$ contains at least $4$ homologies of order $8$, but since the remaining elements of order $8$ $\sigma_{1},\ldots,\sigma_{12}$ are such that $\sigma_{i}^{2}\in SL(2,3)$ is of type (B1) for every $i=1,\ldots,12$ then they cannot be of type (A). The elements of order $4$ split into 2 conjugacy classes: 2 elements are of type (A) and belong to $Z(G)$, while the remaining are of type (B1). From the Riemann-Hurwitz formula

[TABLE]

if $q\equiv_{12}7$, and

[TABLE]

if $q\equiv_{12}11$. The claim now follows by direct computation.

∎

Remark 4.12*.*

If $\alpha\in Z({\mathcal{M}_{q}})$, then the genera $\bar{g}$ given in Proposition 4.11 are actually obtained for some $G\leq PGU(3,q)$. If $\alpha\notin Z({\mathcal{M}_{q}})$, then $G$ as in the hypothesis of Proposition 4.11 may exist or not. For instance, tests with MAGMA show that $PGU(3,11)$ has subgroups $SmallGroup(48,33)$ and $SmallGroup(96,67)$, $PGU(3,13)$ has a subgroup $SmallGroup(48,29)$, and $PGU(3,19)$ has a subgroup $SmallGroup(48,33)$.

Proposition 4.13**.**

If there exists a subgroup $G=K\rtimes C_{m}$ of ${\mathcal{M}_{q}}$ where $K\cong SmallGroup(48,28)$, $p\geq 5$, $q^{2}\equiv 1\mod 16$, $\langle\alpha\rangle=C_{m}$ and $m\mid(q+1)$ such that $\alpha$ is of type (A); then the genus $\bar{g}$ of the quotient curve ${\mathcal{H}}_{q}/G$ is given by one of the following values

[TABLE]

where $D=(m,3)$.

Proof.

Table 3 summarizes the values of $i(\sigma)$, for $\sigma\in K$ according to the congruence of $q$ modulo $24$.

Assume that $\alpha$ and $K$ commute, that is, $\alpha\in Z({\mathcal{M}_{q}})$. Since $|K|$ is even, then $m$ must be odd since otherwise $K$ and $C_{m}$ cannot be disjoint. Assume that $q\equiv_{24}1,13$. Then from the Riemann-Hurwitz formula

[TABLE]

[TABLE]

Assume that $q\equiv_{24}5,17$. As before we observe that we can have $\beta\alpha\in G$ of type (A) if and only if $o(\beta)=o(\alpha)=3$. Then from the Riemann-Hurwitz formula

[TABLE]

[TABLE]

where $D=(3,m)$. Assume that $q\equiv_{24}7,19$. Then from the Riemann-Hurwitz formula

[TABLE]

[TABLE]

Assume that $q\equiv_{24}11,23$. As before we observe that we can have $\beta\alpha\in G$ of type (A) if and only if $o(\beta)=o(\alpha)=3$. Then from the Riemann-Hurwitz formula

[TABLE]

[TABLE]

where as before $D=(3,m)$.

We now assume that $\alpha\not\in Z({\mathcal{M}_{q}})$. We note that $K$ has a subgroup $H\cong SL(2,3)$ which is characteristic, as it is the unique subgroup of $K$ having order equal to $24$. Thus, $H$ is normal in $G$ and $\langle H,\alpha\rangle=H\rtimes C_{m}\cong SL(2,3)\rtimes C_{m}$ because $C_{m}$ and $H$ are disjoint. If $m$ is odd then $\alpha$ commutes with $H$ from [25, Proposition 1.2], and hence $\alpha\in Z({\mathcal{M}_{q}})$, a contradiction. Thus we can assume that $m$ is even. We note that if $m=2^{h}\tilde{m}$, where $\tilde{m}$ is odd, then from [25, Proposition 1.2] $C_{\tilde{m}}<C_{m}$ is a group of elements of type (A) belonging to $Z({\mathcal{M}_{q}})$, a contradiction to $\alpha\not\in Z({\mathcal{M}_{q}})$. In particular this shows that $G=K\rtimes C_{2^{h}}$ with $h\geq 1$. If $q\equiv_{24}1,13,5,17$ then $2\mid(q+1)$ but $4\nmid(q+1)$, thus $G=K\rtimes C_{2}$. Suppose that $q\equiv_{24}7,19$ and that $o(\alpha)>4$. Then $\alpha^{\prime}$ acts on a set of $4$ triangle of type (B2) which are the sets of the fixed points of the elements of order $3$ contained in $K$. Since $\alpha^{\prime}$ cannot fix any of these points, we have a contradiction. Thus, if $q\equiv_{24}7,19$, then either $G=K\rtimes C_{2}$ or $G=K\rtimes C_{4}$. Assume that $q\equiv_{24}11,23$. Arguing as before, since $\alpha$ acts on a set of $4$ triangles of type (B1) which are the fixed points of the elements of order $3$, we get that either $m=2$ or $m=4$.

By direct checking with MAGMA, there are no groups of order $96$ of the form $G=K\rtimes C_{2}$, then both the cases $m=2$ and $m=4$ are impossible. ∎

Remark 4.14*.*

Note that groups $G$ with $\bar{g}$ as in the thesis of Proposition 4.13 do exist for any $q$; conversely, the genera of ${\mathcal{H}}_{q}/G$ with $G$ satisfying the hypothesis of Proposition 4.13 are all given in this proposition.

Proposition 4.15**.**

Let $G=Q_{8}\rtimes C_{m}\leq PGU(3,q)$, where $Q_{8}$ is the quaternion group of order $8$ and $C_{m}=\langle\alpha\rangle$ with $m\mid(q+1)$, $\alpha$ of type (A). Then ${\mathcal{H}}_{q}/G$ has genus

[TABLE]

Proof.

Suppose $\alpha\in Z({\mathcal{M}_{q}})$, so that $G=Q_{8}\times C_{m}$ and $m$ is odd. If $4$ divides $q-1$ then from the Riemann-Hurwitz formula,

[TABLE]

if $4$ divides $q+1$ then

[TABLE]

Suppose $\alpha\not\in Z({\mathcal{M}_{q}})$. Assume that $m=2$. By direct checking in MAGMA the only possibilities are $G\cong SmallGroup(16,8)$ or $G\cong SmallGroup(16,13)$. Let $G\cong SmallGroup(16,8)$. Then

[TABLE]

if $4$ divides $q-1$, and

[TABLE]

if $8$ divides $q+1$; the case $4\mid(q+1)$ and $8\nmid(q+1)$ is not possibile by direct inspection of $SmallGroup(16,8)$. Let $G\cong SmallGroup(16,13)$. Then

[TABLE]

if $8$ divides $q-1$ (the case $4\mid(q-1)$ and $8\nmid(q-1)$ as $\bar{g}$ is integer),

[TABLE]

if $4$ divides $q+1$. From now on, $m>2$. Assume that $\alpha$ fixes pointwise the fixed points of an element $\beta\in Q_{8}$ of order $4$. Then $m=4$. This case is not possible; in fact, by direct inspection with MAGMA, all groups of type $Q_{8}\rtimes C_{4}$ have $3$ normal involutions. Such a groups in $PGU(3,q)$ are abelian, a contradiction. The remaining cases require that $\alpha$ acts with long orbits on a set of $6$ points and hence either $m=3$ or $m=6$. Suppose $m=3$. There exists just one group $SmallGroup(24,3)$ of type $Q_{8}\rtimes C_{3}$, and

[TABLE]

if $4$ divides $q-1$, and

[TABLE]

if $4$ divides $q+1$. Suppose $m=6$. By direct checking with MAGMA a group of type $Q_{8}\rtimes C_{6}$ does not exist. ∎

Remark 4.16*.*

If $\alpha\in Z({\mathcal{M}_{q}})$ then $\bar{g}$ as in Proposition 4.15 exists; other instances are the following: $PGU(3,11)$ contains $SmallGroup(16,13)$ and $SmallGroup(24,3)$, $PGU(3,13)$ contains $SmallGroup(16,8)$, $PGU(3,17)$ contains $SmallGroup(24,3)$.

Proposition 4.17**.**

If there exists a subgroup $G=Dic(n)\rtimes C_{m}$ of ${\mathcal{M}_{q}}$, $n>2$, where $Dic(n)$ is the dicyclic group $Dic(n)=\langle a,x\mid a^{2n}=1,\ x^{2}=a^{n},\ x^{-1}ax=a^{-1}\rangle$ of order $4n$ for $n\mid\frac{q\pm 1}{2}$, $\langle\alpha\rangle=C_{m}$ and $m\mid(q+1)$ such that $\alpha$ is of type (A); then the genus $\bar{g}$ of the quotient curve ${\mathcal{H}}_{q}/G$ is given by one of the following values.

• •

If $q\equiv_{4}1$:

[TABLE]

• •

If $q\equiv_{4}3$:

[TABLE]

where $D=\sum_{d\mid(m,n)}\varphi(d)$, and $\varphi$ denote the Euler totient function.

Proof.

We observe that if $n\neq 2$, then $C_{2n}=\langle a\rangle$ is a characteristic subgroup of $Dic(n)$ as it is the unique cyclic group of $Dic(n)$ of order $2n$. In fact the following is the complete list of the elements of $Dic(n)$,

• •

its unique central element, which is the involution $a^{n}=x^{2}$,

• •

the nontrivial elements $\sigma\in\langle a\rangle\setminus\{a^{n}\}$. They are $2n-2$, and the conjugacy class of each of them has length 2 (contains just $\sigma$ and $\sigma^{-1}$). Thus, there are $n-1$ conjugacy classes of elements of type $\sigma$.

• •

the elements $\beta\in Dic(n)\setminus\langle a\rangle$ are divided into two conjugacy classes:

1. (1)

the conjugacy class of $x$, which contains the elements of order $4$ of type $a^{2k}x$ for $k\geq 0$, 2. (2)

the conjugacy class of $ax$, containing the remain elements of type $a^{2k+1}x$ for $k\geq 0$. Also these elements are of order $4$, as $(ax)^{2}=x(x^{-1}ax)ax=xa^{-1}ax=x^{2}$.

This observation proves that $Dic(n)$ has one element of order $2$, $2n-2$ nontrivial elements of order which divides $2n$ and is greater than $2$, and $2n$ elements of order $4$. Since $\alpha$ acts by conjugation on the set of the cyclic subgroups of $Dic(n)$ of order $4$, and an element of order $4$ is either of type (B1) or (B2) from Theorem 2.3, we have that $\alpha$ acts on a set of $n+1$ triangles if $n$ is even, while it acts on a set of $n$ triangles if $n$ is odd. Assume that $\alpha\in Z({\mathcal{M}_{q}})$. Since $2\mid|Dic(n)|$, $m$ must be odd.

Case 1: $\bf q\equiv_{4}1$, $\bf n\mid(q-1)/2$

All the nontrivial elements of $Dic(n)\setminus\{a^{n}\}$ are of type (B2), while $a^{n}$ is of type (A). From the Riemann-Hurwitz formula

[TABLE]

[TABLE]

Case 2: $\bf q\equiv_{4}1$, $\bf n\mid(q+1)/2$

Here all the nontrivial elements of $C_{2n}\setminus\{a^{n}\}$ are of type (B1), $a^{n}$ is of type (A) and the remaining $2n$ are of type (B2). From the Riemann-Hurwitz formula

[TABLE]

[TABLE]

Case 3: $\bf q\equiv_{4}3$, $\bf n\mid(q-1)/2$

Since $a$ is of type (B2), $n$ must be odd. In fact if $n$ is even, then $C_{2n}$ contains elements of order $4$, and then of type (B1), which are powers of an element of type (B2), a contradiction. Thus, the nontrivial elements of $C_{2n}\setminus\{a^{n}\}$ are of type (B2), the $2n$ elements of order $4$ are of type (B1) and $a^{n}$ is of type (A). From the Riemann-Hurwitz formula

[TABLE]

[TABLE]

Case 4: $\bf q\equiv_{4}3$, $\bf n\mid(q+1)/2$

In this case the elements of $Dic(n)\setminus\{a^{n}\}$ are all of type (B1), while $a^{n}$ is of type (A). From the Riemann-Hurwitz formula

[TABLE]

[TABLE]

We now assume that $\alpha\not\in Z({\mathcal{M}_{q}})$.

Case 1: $\bf q\equiv_{4}1$, $\bf n\mid(q-1)/2$

Since $C_{2n}$ is normal in $G$, and its generator $a$ is of type (B2) we have that $o(\alpha)=2$, since homologies have long orbits outside its center and axis and $\alpha\not\in Z({\mathcal{M}_{q}})$. Furthermore by direct checking in matrix representation, as in the proof of Proposition 4.5, every element $a^{k}\alpha$ where $k\neq n$ is of order $2$. Looking at $\beta\alpha$ for $\beta\not\in C_{2n}$ we observe that $\beta$ and $\alpha$ has the same orbit of length $2$ given by the fixed points of $\alpha$ which is of type (B2). Thus, these two points are fixed by $\beta\alpha$. Since $\beta\alpha$ cannot be a homology, as the fixed points of $\beta$ are not fixed by $\beta\alpha$ we get that $\beta\alpha$ is of type (B2) as well. From the Riemann-Hurwitz formula

[TABLE]

Case 2: $\bf q\equiv_{4}1$, $\bf n\mid(q+1)/2$

Since $C_{2n}$ is normal in $G$, and its generator $a$ is of type (B1), we have that either $o(\alpha)=2$ or $\alpha$ and $C_{2n}$ commute, since homologies have long orbits apart from their center and axis and $\alpha\not\in Z({\mathcal{M}_{q}})$.

Assume that $o(\alpha)=2$ and that $\alpha$ and $C_{2n}$ do not commute. Let $\mathcal{H}_{q}$ have equation (2.1). Up to conjugation, $a$ is the diagonal matrix $[\lambda,\lambda^{i},1]$, with $\lambda$ a $2n$-th primitive root of unity, and

[TABLE]

with $A^{q+1}=1$. By direct checking, $x^{-1}ax=a^{-1}$ if and only if $i=-1$. Moreover, $\alpha x\alpha\in Dic_{n}$ if and only if $ord(A)\mid 2n$. As $(a^{k}\alpha)^{2}=1$, every element of type $a^{k}\alpha$ is of type (A). Also, $a^{k}x\alpha=[\lambda^{k}A,-(\lambda^{k}A)^{-1},1]$ is of type (A) if $\lambda^{k}=\pm A^{-1}$, while $a^{k}x\alpha$ is of type (B1) otherwise. From the Riemann-Hurwitz formula,

[TABLE]

Assume that $\alpha$ and $C_{2n}$ commute. As before, $\mathcal{H}_{q}$ has equation (2.1), $a={\rm diag}[\lambda,\lambda^{-1},1]$. Also, $\alpha={\rm diag}[1,\mu,1]$. Since $\alpha$ acts on the elements of order $4$ in $Dic_{n}\setminus C_{2n}$, we have $ord(\mu)\mid 2n$. Thus $\mu=\lambda^{j}$ for some $j$; hence, $a^{j}\alpha^{2}={\rm diag}[\mu,\mu,1]\in Z(\mathcal{M}_{q})$ and $G=Dic_{n}\times\langle a^{j}\alpha^{2}\rangle$. This case has already been considered.

Case 3: $\bf q\equiv_{4}3$, $\bf n\mid(q-1)/2$

Suppose that $\alpha$ and $C_{2n}$ do not commute. Let $\mathcal{H}_{q}$ have equation (2.2); up to conjugation, $a=[\lambda^{2},\lambda,1]$ with $\lambda$ a $2n$-th primitive root of unity,

[TABLE]

where $A^{q-1}=-1$. By direct checking, $\alpha x\alpha\in Dic_{n}$ if and only if $\lambda^{j}=-A^{-2}$ for some $j$, which implies $A^{2n}=1$. Since $2n\mid(q-1)$, this is a contradiction to $A^{q-1}=-1$. Thus, $\alpha$ and $C_{2n}$ commute. This implies that $\alpha\in Z(\mathcal{M}_{q})$, and this case has already been considered. ∎

Remark 4.18*.*

From the proof of the previous proposition, we note that all the integers $\bar{g}$ given in Proposition 4.17 actually occur as genera of some quotient ${\mathcal{H}}_{q}/G$; viceversa, if $G$ satisfies the hypothesis, then the genus of ${\mathcal{H}}_{q}/G$ is given in the thesis of Proposition 4.17.

5. Further results

Proposition 5.1**.**

Let $p\geq 2$ and $G\leq PGU(3,q)$ with $G\cong PGU(3,\bar{q})$, where $q=p^{h}$, $\bar{q}=p^{k}$, $k$ divides $h$, and $h/k$ is odd. Then the quotient curve ${\mathcal{H}}_{q}/G$ has genus

[TABLE]

where

[TABLE]

[TABLE]

with

[TABLE]

Proof.

Up to conjugation $G$ is the group of all $\sigma\in PGU(3,q)$ such that $\sigma$ is defined over $\mathbb{F}_{{\bar{q}}^{2}}$. First, we classify the elements of $PGU(3,\bar{q})$ seen as the automorphism group of a Hermitian curve ${\mathcal{H}}_{\bar{q}}$, using the order statistics of $PGU(3,{\bar{q}})$ and Lemma 2.3.

1. (1)

There are exactly $(\bar{q}-1)(\bar{q}^{3}+1)$ elements of type (C). In fact, for each $P\in{\mathcal{H}}_{\bar{q}}(\mathbb{F}_{{\bar{q}}^{2}})$, there exist exactly ${\bar{q}}-1$ elations in $PGU(3,{\bar{q}})$ with center $P$. 2. (2)

There are exactly $(\bar{q}^{3}-\bar{q})(\bar{q}^{3}+1)$ elements of type (D). In fact, they are the $p$-elements of $PGU(3,{\bar{q}})$ which are not of type (C). 3. (3)

There are exactly $\bar{q}({\bar{q}}^{4}-{\bar{q}}^{3}+{\bar{q}}^{2})$ elements of type (A). In fact, for each $P\in PG(2,{\bar{q}}^{2})\setminus{\mathcal{H}}_{\bar{q}}$ there exist exactly ${\bar{q}}$ homologies in $PGU(3,{\bar{q}})$ with center $P$. 4. (4)

There are exactly $({\bar{q}}^{2}-{\bar{q}}-2)\frac{({\bar{q}}^{3}+1){\bar{q}}^{3}}{2}$ elements of type (B2). In fact, for each couple $\{P,Q\}\subset{\mathcal{H}}_{\bar{q}}(\mathbb{F}_{{\bar{q}}^{2}})$ there exist exactly ${\bar{q}}^{2}-2$ nontrivial elements of $PGU(3,{\bar{q}})$ fixing $P$ and $Q$, and ${\bar{q}}$ of them are homologies with axis $PQ$. 5. (5)

There are exactly $({\bar{q}}^{2}-{\bar{q}})\frac{{\bar{q}}^{6}+{\bar{q}}^{5}-{\bar{q}}^{4}-{\bar{q}}^{3}}{3}$ elements of type (B3). In fact, any point $P\in{\mathcal{H}}_{{\bar{q}}^{2}}(\mathbb{F}_{{\bar{q}}^{6}})\setminus{\mathcal{H}}_{{\bar{q}}}(\mathbb{F}_{{\bar{q}}^{2}})$ determines a unique triangle $\{P,\Phi_{{\bar{q}}^{2}}(P),\Phi_{{\bar{q}}^{2}}^{2}(P)\}\subset{\mathcal{H}}_{{\bar{q}}}(\mathbb{F}_{{\bar{q}}^{6}})$ fixed by a Singer subgroup of order ${\bar{q}}^{2}-{\bar{q}}+1$; see the proof of Lemma 2.3 given in [22]. 6. (6)

There are exactly $({\bar{q}}-1){\bar{q}}({\bar{q}}^{3}+1){\bar{q}}^{2}$ elements of type (E). In fact, consider a couple $\{P,Q\}$ with $P\in{\mathcal{H}}_{\bar{q}}(\mathbb{F}_{{\bar{q}}^{2}})$ and $Q\in PG(2,{\bar{q}}^{2})\cap\ell_{P}$, where $\ell_{P}$ is the tangent line to ${\mathcal{H}}_{\bar{q}}$ at $P$. Any element of type (E) fixing $P$ and $Q$ is uniquely obtained as the product of an elation of center $P$ and a homology of center $Q$; thus there exist exaclty $({\bar{q}}-1){\bar{q}}$ such elements. 7. (7)

The remaining ${\bar{q}}^{4}({\bar{q}}-1)^{2}({\bar{q}}^{2}-{\bar{q}}+1)/6$ nontrivial elements are of type (B1).

Now we describe the elements in each class (1) - (7) according to their geometry with respect to ${\mathcal{H}}_{q}$, using Lemma 2.3.

• (i)

The elements in class (1) are of type (C). In fact, let $\bar{S}$ be one of the ${\bar{q}}^{3}+1$ Sylow $p$-subgroups of $PGU(3,{\bar{q}})$ and $S$ be the Sylow $p$-subgroup of $PGU(3,q)$ containing $\bar{S}$. Note that $S$ is a trivial intersection set, since ${\mathcal{H}}_{q}$ has zero $p$-rank; see [17, Theorem 11.133]. Consider the explicit representation of $S$ given in [13, Section 3], where ${\mathcal{H}}_{q}$ has norm-trace equation and $S$ fixes the point at infinity. By direct computation, there are ${\bar{q}}$ elements of $\bar{S}$ in the center of $S$. Thus, $\bar{S}$ has exactly ${\bar{q}}-1$ elements of type (C); see [13, Equation (2.12)].

• (ii)

The elements in class (2) are of type (D). In fact, with $\bar{S}$ as in Case (i), the claim follows from Case (i) counting the remaining nontrivial elements of $\bar{S}$.

• (iii)

Let $\sigma\in PGU(3,{\bar{q}})$ be in class (3) or (4). Then $\sigma$ is contained in the pointwise stabilizer $D\cong C_{{\bar{q}}+1}\times C_{{\bar{q}}+1}$ of a self-polar triangle with respect to ${\mathcal{H}}_{\bar{q}}$; see Theorem 2.1. Let $C_{q+1}\times C_{q+1}\leq PGU(3,q)$ be the poitwise stabilizer of a self-polar triangle $T$ with respect to ${\mathcal{H}}_{q}$, such that $D\leq C_{q+1}\times C_{q+1}$. Up to conjugation, ${\mathcal{H}}_{q}$ has Fermat equation (2.1) and $T$ is the fundamental triangle, so that

[TABLE]

By direct computation, $D$ contains $3{\bar{q}}$ elements of type (A) and ${\bar{q}}^{2}-{\bar{q}}$ elements of type (B1). Since $PGU(3,{\bar{q}})$ is transitive on $PG(2,{\bar{q}}^{2})\setminus{\mathcal{H}}_{q}$, there are exactly $\frac{|PGU(3,{\bar{q}})|}{6|D|}$ self-polar triangles $T^{\prime}$ with respect to ${\mathcal{H}}_{\bar{q}}$, whose pointwise stabilizer $D^{\prime}$ is conjugated to $D$ under $PGU(3,{\bar{q}})$. Note that $D$ and $D^{\prime}$ intersect non-trivially if and only if $T$ and $T^{\prime}$ have a vertex $P$ in common; in this case, $(D\cap D^{\prime})\setminus\{id\}$ is made by ${\bar{q}}$ homologies with center $P$. The number of points in $PG(2,{\bar{q}}^{2})\setminus{\mathcal{H}}_{q}$ lying on the polar of $P$ is ${\bar{q}}^{2}-{\bar{q}}$; hence, the number of $D^{\prime}$ which intersect $D$ non-trivially is $({\bar{q}}^{2}-{\bar{q}})/2$. Therefore, by direct computation, the number of element of type (A) in $PGU(3,{\bar{q}})$ is exactly the number ${\bar{q}}({\bar{q}}^{4}-{\bar{q}}^{3}+{\bar{q}}^{2})$ of elements in class (3), and the remaining ${\bar{q}}^{4}({\bar{q}}-1)^{2}({\bar{q}}^{2}-{\bar{q}}+1)/6$ elements of the subgroups of $PGU(3,{\bar{q}})$ conjugated to $D$ are of type (B1) and in class (7).

• (iv)

Let $\sigma\in PGU(3,{\bar{q}})$ be in class (4). Since the order $o(\sigma)$ of $\sigma$ divides ${\bar{q}}^{2}-1$ but not ${\bar{q}}+1$, we have that $o(\sigma)$ divides $q^{2}-1$ but not $q+1$, as $q$ is an odd power of ${\bar{q}}$. Therefore $\sigma$ is of type (B2).

• (v)

Let $\sigma\in PGU(3,{\bar{q}})$ be in class (6). Since the order of $\sigma$ is $p\cdot d$ where $d>1$ and $p\nmid d$, $\sigma$ is of type (E).

• (vi)

Let $\sigma\in PGU(3,{\bar{q}})$ be in class (5). By direct checking, the order $o(\sigma)$ of $\sigma$ divides either $q^{2}-q+1$ or $q+1$. Assume that ${\bar{q}}^{2}-{\bar{q}}+1\neq 3$. If ${\bar{q}}^{2}-{\bar{q}}+1\mid(q^{2}-q+1)$ then every $\sigma$ in class (5) is of type (B3), by Lemma 2.3. If ${\bar{q}}^{2}-{\bar{q}}+1\mid(q+1)$ then $\sigma$ is either of type (A) or (B1). We note that $\sigma$ is contained in the maximal subgroup (iv) of $PGU(3,{\bar{q}})$ in Theorem 2.1, which is a semidirect product $C_{{\bar{q}}^{2}-{\bar{q}}+1}\rtimes C_{3}$, where $\sigma\in C_{{\bar{q}}^{2}-{\bar{q}}+1}$ and $C_{3}$ does not commute with any subgroup of $C_{{\bar{q}}^{2}-{\bar{q}}+1}$. Therefore $\sigma$ cannot be of type (A), because otherwise every element which normalizes $\sigma$ should commute with $\sigma$. Then $\sigma$ of type (B1). Now assume ${\bar{q}}^{2}-{\bar{q}}+1=3$, that is ${\bar{q}}=2$. Let $S$ be a Singer subgroup of $PGU(3,q)$ containing $\sigma$ as in case (iv) of Theorem 2.1. Using the explicit description of $S$ given in [7, 8], it is easily seen that $S\cap PGU(3,{\bar{q}})$ is cyclic of order $3$. Then $\sigma$ is of type (B3).

From Theorem 2.4, the degree $\Delta$ of the different divisor of the cover ${\mathcal{H}}_{\bar{q}}\to{\mathcal{H}}_{\bar{q}}/G$ is

[TABLE]

[TABLE]

where $\delta=3$ if $({\bar{q}}^{2}-{\bar{q}}+1)\mid(q^{2}-q+1)$, and $\delta=0$ if $({\bar{q}}^{2}-{\bar{q}}+1)\mid(q+1)$ and ${\bar{q}}\neq 2$. From the Riemann-Hurwitz genus formula, the claim follows. ∎

Proposition 5.2**.**

Let $p\geq 2$, $2\neq n\mid(q+1)$, $n$ prime, $3\mid(q-1)$, $3\mid(n-1)$. Then there exists a subgroup $G=C_{n}\rtimes C_{3}\leq PGU(3,q)$ such that the genus $\bar{g}$ of the quotient curve ${\mathcal{H}}_{q}/G$ is

[TABLE]

Proof.

Let ${\mathcal{H}}_{q}:X^{q+1}+Y^{q+1}+Z^{q+1}=0$, and consider the following automorphisms of ${\mathcal{H}}_{q}$

[TABLE]

where $\lambda^{n}=1$ and

[TABLE]

where $A^{q+1}=B^{q+1}=1$. Since $n$ is prime and $n\equiv 1\pmod{3}$, we have $\gamma:=\beta^{-1}\alpha\beta=\alpha^{n-i}$. Thus, $G:=\langle\alpha,\beta\rangle$ is a nontrivial semidirect product $C_{n}\rtimes C_{3}$. Also, the elements of $C_{n}$ are of type (B1), while the elements of $G\setminus C_{n}$ have order $3$ and are of type (B2). From the Riemann-Hurwitz formula,

[TABLE]

∎

Proposition 5.3**.**

Let $p>3$. Then there exists a subgroup $G\leq PGU(3,q)$ such that $G$ is isomorphic to the alternating group $A_{4}$ and the genus $\bar{g}$ of ${\mathcal{H}}_{q}/G$ is

[TABLE]

Proof.

Define $G=\langle\alpha,\beta,\gamma\rangle$ with $\alpha:(X,Y,Z)\mapsto(-X,Y,Z)$, $\beta:(X,Y,Z)\mapsto(X,-Y,Z)$, $\alpha:(X,Y,Z)\mapsto(Y,Z,X)$. Then $G\cong A_{4}$, and from the Riemann-Hurwitz formula the claim follows. ∎

Proposition 5.4**.**

Let $p=3$. Then there exists a subgroup $G\leq PGU(3,q)$ such that $G$ is isomorphic to the symmetric group $S_{3}$ and the genus $\bar{g}$ of ${\mathcal{H}}_{q}/G$ is

[TABLE]

Proof.

Let ${\mathcal{H}}_{q}$ have equation (2.1). Let $M\cong(C_{q+1}\times C_{q+1})\rtimes K$, with $K\cong S_{3}$, be the maximal subgroup of $PGU(3,q)$ fixing the fundamental triangle $T=\{P_{1},P_{2},P_{3}\}$. Assume $3\mid(q+1)$. Since $M$ contains a unique subgroup $C_{3}$ generated by an element $\alpha$ of type (B1), we have $G=\langle\alpha,\beta\rangle\cong S_{3}$, where $\beta$ is any involution of $M$. Then from the Riemann-Hurwitz formula

[TABLE]

Assume $3\mid(q-1)$ and take $G=K$. Then from the Riemann-Hurwitz formula

[TABLE]

∎

We provide a model for a quotient curve ${\mathcal{H}}_{q}/\langle\sigma\rangle$ where $\sigma$ is of type (E).

Proposition 5.5**.**

For $q=p^{h}$ let $\sigma\in PGU(3,q)$ be of type (E) and of order $pd$. Choose $\lambda\in\overline{\mathbb{F}}_{q}$ such that $\lambda^{h}=-1$. Then a plane model for the quotient curve ${\mathcal{H}}_{q}/\langle\sigma\rangle$ is given by

[TABLE]

Proof.

Let $\overline{\mathbb{F}}_{q}({\mathcal{H}}_{q})=\overline{\mathbb{F}}_{q}(x,y)$ with $y^{q+1}=x^{q}+x$. Up to conjugation $\sigma:(x,y)\mapsto(x+c,ay)$ where $a$ is a $d$-th primitive root of unity and $c^{p}=\lambda c$. Define $\xi=y^{d}$ and $\eta=x^{p}-\lambda x$. Clearly $\overline{\mathbb{F}}_{q}(\xi,\eta)\subseteq\overline{\mathbb{F}}_{q}({\mathcal{H}}_{q})^{\langle\sigma\rangle}$, where $\overline{\mathbb{F}}_{q}({\mathcal{H}}_{q})^{\langle\sigma\rangle}$ denotes the subfield of $\overline{\mathbb{F}}_{q}({\mathcal{H}}_{q})$ which is fixed by $\langle\sigma\rangle$. Since $[\overline{\mathbb{F}}_{q}(y,\eta):\overline{\mathbb{F}}_{q}(\xi,\eta)]=d$ and $[\overline{\mathbb{F}}_{q}(\xi,x):\overline{\mathbb{F}}_{q}(\xi,\eta)]=p$, the compositum $\overline{\mathbb{F}}_{q}(x,y)$ has degree $dp$ over $\overline{\mathbb{F}}_{q}(\xi,\eta)$ by Abhyankar’s Lemma [23, Theorem 3.9.1] and hence $\overline{\mathbb{F}}_{q}({\mathcal{H}}_{q})^{\langle\sigma\rangle}=\overline{\mathbb{F}}_{q}(\xi,\eta)$. Consider the conventional $p$-associate polynomials $x-\lambda$ and $x^{h}-\lambda^{h}$ of $x^{p}-\lambda x$ and $x^{q}+x$; see [20, Definition 3.58]. Since $(x^{h}-\lambda^{h})/(x-\lambda)=\sum_{i=1}^{h}\lambda^{i-1}x^{h-i}$, we have by [20, Lemma 3.59] that

[TABLE]

The equation $\xi^{(q+1)/d}=\sum_{i=1}^{h}\lambda^{i-1}\eta^{q/p^{i}}$ is absolutely irreducible as it defines the Kummer extension $\overline{\mathbb{F}}_{q}(\xi,\eta):\overline{\mathbb{F}}_{q}(\eta)$ of degree $(q+1)/d$. ∎

Remark 5.6*.*

Let $\mathcal{X}$ be the quotient curve ${\mathcal{H}}_{q}/\langle\sigma\rangle$ for some nontrivial $\sigma\in PGU(3,q)$, $q=p^{h}$. Then one of the following cases occurs:

1. (1)

$o(\sigma)=p$, $g(\mathcal{X})=p^{h-1}(q-p)/2$, and

[TABLE]

with $\omega^{q-1}=-1$; see [8, Theorem 2.1 (II) (1)]. 2. (2)

$p\geq 3$, $o(\sigma)=p$, $g(\mathcal{X})=p^{h-1}(q-1)/2$, and

[TABLE]

see [8, Theorem 2.1 (II) (2)]. 3. (3)

$o(\sigma)=pd$ with $1<d\mid(q+1)$, $g(\mathcal{X})=\left(\frac{q+1}{d}-1\right)\left(p^{h-1}-1\right)/2$, and

[TABLE]

see [13, Theorem 4.4] and Proposition 5.5. 4. (4)

$o(\sigma)=d\mid(q^{2}-q+1)$, $g(\mathcal{X})=\frac{1}{2}\left(\frac{q^{2}-q+1}{d}-1\right)$, and

[TABLE]

where $F(X,Y,Z)$ is defined in [7, Remark 5.5]; see [7, Theorem 5.1]. 5. (5)

$o(\sigma)=d\mid(q^{2}-1)$, $g(\mathcal{X})=\frac{1}{2d}[q+1-\gcd(d,q+1)](q-1)$, and

[TABLE]

see [13, Corollary 4.9 and Example 6.3]. 6. (6)

$o(\sigma)=d\mid(q+1)$, $\sigma$ is of type (B1), and $g(\mathcal{X})=1+\frac{1}{2d}(q+1)(q+1-r_{1}-r_{2}-r_{3})$ for some divisors $r_{1},r_{2},r_{3}$ of $q+1$; see [13, Theorem 5.8]. 7. (7)

$p=2$, $o(\sigma)=4$, $\sigma$ is of type (D), and $g(\mathcal{X})=(q^{2}-2q)/8$; see [13, Theorem 3.3].

Partial results on models for $\mathcal{X}$ in Case (6) are given in [13, Example 6.4] and [14, Equation (3.1)]; yet, a model of $\mathcal{X}$ is still unknown for a general $\sigma$ of type (B1). In case (7), a model for $\mathcal{X}$ is unknown.

6. Spectrum of genera of quotients of ${\mathcal{H}}_{q}$, $q\leq 29$

In this section the complete spectrum $\Sigma_{q}$ of genera of $\mathbb{F}_{q^{2}}$-maximal curves which are Galois subcovers of the Hermitian curve ${\mathcal{H}}_{q}$ is determined for all $q\leq 29$.

The proof relies on the results of [22]. A case-by-case analysis of all integers $g$ with $1<g\leq g({\mathcal{H}}_{q})$ is combined with the classical bounds

[TABLE]

This leads us to look inside the structure of the groups $G$ satisfying (6.1) and compute the genus of ${\mathcal{H}}_{q}/G$, for $g>1$. For each $g>1$ in $\Sigma_{q}$, the following tables provide a classification of the groups $G$ for which ${\mathcal{H}}_{q}/G$ has genus $g$.

7. New genera for maximal curves

The results of the previous sections provide new genera for maximal curves over finite fields. To exemplify this fact, we collect in this section new genera for small values of $q$.

Remark 7.1*.*

Tables 12 and 13 partially answer to [3, Remark 4.4]. In particular, we have obtained a $\mathbb{F}_{13^{2}}$-maximal curve of genus $4$ from Prop. 5.2, a $\mathbb{F}_{13^{2}}$-maximal curve of genus $5$ from Prop. 5.3, and a $\mathbb{F}_{16^{2}}$-maximal curve of genus $16$ from Prop. 5.4.

Remark 7.2*.*

The value $g=3$ in Table 15 is new in the spectrum of genera of $\mathbb{F}_{13^{2}}$-maximal curves. In fact this value is given in [10] as an application of [13, Proposition 4.6], but the actual value which provided by [13, Proposition 4.6] is $g=0$. Moreover, it can be checked that if $g({\mathcal{H}}_{13}/G)=3$ then $G\leq\mathcal{M}_{13}$.

Remark 7.3*.*

The genera $g=3$, $g=8$ and $g=22$ provided in Table 14 for $\mathbb{F}_{25^{2}}$-maximal curves are new with respect to [10].