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EXTRACTA

MATHEMATICAE

Volumen 33, Número 2, 2018

instituto de investigación de matemáticas de la

universidad de extremadura

EXTRACTA MATHEMATICAE

Vol. 37, Num. 2 (2022), 195 – 210

doi:10.17398/2605-5686.37.2.195

Available online July 4, 2022

Genus zero of projective symplectic groups

H.M. Mohammed Salih, Rezhna M. Hussein

Department of Mathematics, Faculty of Science, Soran University

Kawa St. Soran, Erbil, Iraq

havalmahmood07@gmail.com , rezhnarwandz@gmail.com

Received January 17, 2022 Presented by A. Turull
Accepted May 23, 2022

Abstract: A transitive subgroup G ≤ SN is called a genus zero group if there exist non identity
elements x1, . . . ,xr ∈ G satisfying G = 〈x1,x2, . . . ,xr〉,

∏r
i=1 xi = 1 and

∑r
i=1 ind xi = 2N − 2.

The Hurwitz space Hinr (G) is the space of genus zero coverings of the Riemann sphere P1 with r
branch points and the monodromy group G.

In this paper, we assume that G is a finite group with PSp(4,q) ≤ G ≤ Aut(PSp(4,q)) and G acts
on the projective points of 3-dimensional projective geometry PG(3,q), q is a prime power. We
show that G possesses no genus zero group if q > 5. Furthermore, we study the connectedness of

the Hurwitz space Hinr (G) for a given group G and q ≤ 5.

Key words: symplectic group, fixed point, genus zero group.

MSC (2020): 20B15, 20C33.

1. Introduction

A one dimensional compact manifold is called Riemann surface. Topolog-
ically, such surfaces are either spheres or tori which have been glued together.
The number of holes so joined is called the genus. Let F : X → P1 be a mero-
morphic function from a compact connected Riemann surface X of genus
g into the Riemann sphere P1. For every meromorphic function there is a
positive integer N such that all points have exactly N preimages. So every
compact Riemann surface can be made into the branched covering of P1. It is
known that one of the basic strategies of the whole subject of algebraic topol-
ogy is to find methods to reduce topological problem about continuous maps
and spaces into pure algebraic problems about homomorphisms and groups
by using the fundamental group. The points p are called the branch points
of F if |F−1(p)| < N. It is well known that the set of branch points is finite
and it will be denoted by B = {p1, . . . ,pr}. For q ∈ P1 \B, the fundamental
group π1(P1 \B,q) is a free group which is generated by all homotopy classes
of loops γi winding once around the point pi. These loops of generators γi

ISSN: 0213-8743 (print), 2605-5686 (online)

© The author(s) - Released under a Creative Commons Attribution License (CC BY-NC 3.0)

https://doi.org/10.17398/2605-5686.37.2.195
mailto:havalmahmood07@gmail.com
mailto:rezhnarwandz@gmail.com
https://publicaciones.unex.es/index.php/EM
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196 h.m.m. salih, r.m. hussein

are subject to the single relation that γ1 · . . . · γr = 1 in π1(P1 \ B,q). The
explicit and well known construction of Hurwitz shows that a Riemann surface
X with N branching coverings of P1 is defined in the following way: consider
the preimage F−1(q) = {x1, . . . ,xN}, every loop in γ in P1 \B can be lifted
to N paths γ̃1, . . . , γ̃N where γ̃i is the unique path lift of γ and γ̃i(0) = xi for
every i. The endpoints γ̃i(1) also lie over q. That is γ̃i(1) = xσ(i) in F

−1(q)
where σ is a permutation of the indices {1, . . . ,N} and it depends only on γ.
Thus it gives a group homomorphism φ : π1(P1 \B,q) → SN . The image of φ
is called the monodromy group of F and denoted by G = Mon(X,F). Since
X is connected, then G is a transitive subgroup of SN . Thus a group homo-
morphism is determined by choosing N permutations xi = φ(γi), i = 1, . . . ,r
and satisfying the relations

G = 〈x1,x2, . . . ,xr〉, (1)
r∏
i=1

xi = 1, xi ∈ G# = G\{1}, i = 1, . . . ,r, (2)

r∑
i=1

ind xi = 2(N + g − 1), (3)

where ind x = N − orb(x), orb(x) is the number of orbits of the group gener-
ated by x on Ω where |Ω| = N. Equation (3) is called the Riemann Hurwitz
formula. A transitive subgroup G ≤ SN is called a genus g group if there
exist x1, . . . ,xr ∈ G satisfying (1), (2) and (3) and then we call (x1, . . . ,xr)
the genus g system of G. If the action of G on Ω is primitive, we call G a
primitive genus g group and (x1, . . . ,xr) a primitive genus g system.

A group G is said to be almost simple if it contains a non-abelian simple
group S and S ≤ G ≤ Aut(S). In [4], Kong worked on almost simple groups
whose socle is a projective special linear group. Also, she gave a complete list
for some almost simple groups of Lie rank 2 up to ramification type in her
PhD thesis for genus 0, 1 and 2 system. Furthermore, she showed that the
almost simple groups with socle PSL(3,q) do not possess genus low tuples if
q ≥ 16. In [6], Mohammed Salih gave the classification of some almost simple
groups with socle PSL(3,q) up to braid action and diagonal conjugation.

The symplectic group Sp(n,q) is the group of all elements of GL(n,q)
preserving a non-degenerate alternating form; the non degenerate leads to n
being even. The projective symplectic group PSp(n,q) is obtained by from
Sp(n,q) on factoring it by the subgroup of scalar matrices it contains (which
has order at most 2) [1]. In this paper we consider a finite group G with



genus zero of projective symplectic groups 197

PSp(4,q) ≤ G ≤ Aut(PSp(4,q)) and G acts on the projective points of
3-dimensional projective geometry PG(3,q), q is a prime power.

We will now describe the work carried out in this paper. In the second
section we review some basic concepts and results will be used later. In
the third section, we provide some basic facts for computing fixed points
and generating tuples. Finally, we show that G possesses no genus 0 group
if PSp(4,q) ≤ G ≤ Aut(PSp(4,q)) and G acts on the projective points of
3-dimensional projective geometry PG(3,q), q is a prime power and q > 5.
Furthermore, we study the connectedness of the Hurwitz space G if q ≤ 5.

2. Preliminary

We begin by introducing some definitions and stating a few results which
will be needed later. Assume that G is a finite permutation group of degree N.
The signature of the r-tuple x = (x1, . . . ,xr) is the r-tuple d = (d1, . . . ,dr)
where di = o(xi). We assume that di ≤ dj if i ≤ j, because of the braid
action on x. The following result will tell us the tuple x can not generate G,
where G = PGL(4,q) or PSL(4,q) if (ii), (iii) and (iv) below hold. So, setting
A(d) =

∑r
i=1

di−1
di
, we have A(d) ≥ 85

42
.

Proposition 2.1. ([3]) Assume that a group G acts transitively and
faithfully on Ω and |Ω| = N. Let r ≥ 2, G = 〈x1, . . . ,xr〉,

∏r
i=1 xi = 1

and o(xi) = di > 1, i = 1, . . . ,r. Then one of the following holds:

(i)
∑r

i=1
di−1
di
≥ 85

42
;

(ii) r = 4, di = 2 for each i = 1 and G
′′

= 1;

(iii) r = 3 and (up to permutation) (d1,d2,d3) =

(a) (3, 3, 3), (2, 3, 6) or (2, 4, 4) and G
′′

= 1;

(b) (2, 2,d) and G is dihedral;

(c) (2, 3, 3) and G ∼= A4;
(d) (2, 3, 4) and G ∼= S4;
(e) (2, 3, 5) and G ∼= A5;

(iv) r = 2 and G is cyclic.

For a permutation x of the finite set Ω, let Fix(x) denote the fixed points
of x on Ω and f(x) = |Fix(x)| is the number of fixed points of x. Note that
the conjugate elements have the same number of fixed points. The following
result provides a useful connection between fixed points and indices.



198 h.m.m. salih, r.m. hussein

Lemma 2.2. ([3]) If x is a permutation of order d on a set of size N, then
ind x = N − 1

d

∑
y∈〈x〉f(y) where 〈x〉 is the cyclic group generated by x.

The fixed point ratio of x is defined by fpr(x) =
f(x)
N

. The codimension
of the largest eigenspace of a linear transformation ḡ in GL(n,q) is denoted
by v(ḡ). Ω denotes the set of the projective points of projective geometry
PG(n− 1,q) that is the set of 1-dimensional subspaces of vector space over a
finite field GF(q). In this paper we take |Ω| = q

n−1
q−1 and n = 4, so we have

|Ω| = q3 + q2 + q + 1.
The center of GL(n,q) is the set of all scalar matrices and denoted by

Z(GL(n,q)). The projective general linear group and the projective special
linear group are defined by

PGL(n,q) =
GL(n,q)

Z(GL(n,q))
and PSL(n,q) =

SL(n,q)

Z(SL(n,q))

respectively, where Z(SL(n,q)) = SL(n,q)∩Z(GL(n,q)). They act primitively
on Ω.

Let 〈v〉 ∈ Ω be a fixed point of g ∈ PGL(n,q) and let ḡ be an element
in the preimage of g in GL(n,q) that fixes 〈v〉. The fixed points of g are the
1-spaces spanned by eigenvectors of ḡ. So we classify non identity elements in
PGL(4,q) by their fixed points as follows:

Table 1: Number of Fixed points

v(ḡ) Type of eigenspaces of ḡ ∈ GL(n, q)
Number of fixed

points of g ∈ PGL(4, q)
4 no eigenspace 0

3 one 1-dimensional eigenspace 1

3 two 1-dimensional eigenspaces 2

3 three 1-dimensional eigenspaces 3

2 one 2-dimensional eigenspace q + 1

1 one 3-dimensional eigenspace q2 + q + 1

2 one 1-dimensional and one 2-dimensional eigenspaces q + 2

2 one 2-dimensional and one 2-dimensional eigenspaces 2q + 2

1 one 1-dimensional and one 3-dimensional eigenspaces q2 + q + 2

2
one 1-dimensional, one 1-dimensional and

one 2-dimensional eigenspaces
q + 3

According to Table 1, we have two cases. If v(ḡ) = 1, then g fixes q2 +q+ 1
or q2 + q + 2 points. Otherwise, it fixes at most 2q + 2 points. From this fact,



genus zero of projective symplectic groups 199

we will show that there are no genus zero systems for PSL(4,q) and PGL(4,q)
when q > 37.

The following result is an interesting tool to compute β in the next section.

Lemma 2.3. (Scott Bound, [7]) Let G ≤ GL(n,q). If a triple x =
(x1,x2,x3) satisfies G = 〈x1,x2,x3〉 and x1x2x3 = 1, then v(xi) + v(xj) ≥ n
where i 6= j and 1 ≤ i,j ≤ 3. In particular if n ≥ 3 and i 6= j, then v(xi) ≥ 2
or v(xj) ≥ 2.

Lemma 2.4. ([5]) If 1
N

∑r
i=1

∑di−1
j=1

f(xj )
di

< A(d)−2, then d is not a genus
zero system.

3. Existence of genus zero system

Now, we are going to apply Lemma 2.4, to exclude all signatures which
do not satisfy the Riemann Hurwitz formula. As a result, we will obtain
Theorem 3.1.

Let F be the set of elements with q2 + q + 1 or q2 + q + 2 fixed points in
PGL(4,q). So we have

fpr(x) ≤



q2 + q + 2

N
if x ∈ F,

2q + 2

N
if x /∈ F.

Assume that α =
q2+q+2
N

and γ =
2q+2
N

. Combining the Riemann Hurwitz
formula as done in [4], we obtain the following inequality

A(d) ≤
2 + � + β(α−γ)

1 −γ
(4)

where � = −2
N

and β =
∑r

i=1
|〈xi〉#∩F|

di
. If α = γ in inequality (4), then we

obtain the following

A(d) ≤
2 + �

1 −α
(5)

and we have fpr(x) ≤ α. Now bound β for the tuple x = (x1, . . . ,xr).
Following [4], if xi ∈ F , then every power of xi that are non-identity are

also in F , because if any point fixed by xi, then it is also fixed by x
l
i. Therefore,

f(xi) ≤ f(xli). In this situation, there are di − 1 elements in F in 〈xi〉. If
xi /∈ F, then there are φ(di) generators in 〈xi〉 where φ is the Euler’s function.



200 h.m.m. salih, r.m. hussein

All of these generators are not in F either, so there are at most di−φ(di)−1
elements in F in 〈xi〉. We obtain

β 6
∑
xi∈F

di − 1
di

+
∑
xi /∈F

di −φ(di) − 1
di

.

Notice that Lemma 2.3 will tell us for the tuple of length 3, that at most one
element lies in F.

In PGL(4,q) and PSL(4,q), α =
q2+q+2

q3+q2+q+1
and γ =

2q+2
q3+q2+q+1

, � =
−2

q3+q2+q+1
for every genus 0 tuples.

Let q ≥ 16, then using inequality (5), we have A(d) ≤ 32
15

. If r ≥ 4, then
A(d) ≥ A((2, 2, 2, 3)) = 13

6
. But 32

15
< 13

6
. So the number of branch points r

must be 3.
Now we are looking for signatures which satisfy the inequality 85

42
≤ A(d) ≤

32
15

. This leads d only can be (2, 3,d3) with 7 ≤ d3 ≤ 30, (2, 4,d3) with
5 ≤ d3 ≤ 8, (2, 5, 5), (2, 5, 6), (3, 3, 4), (3, 3, 5).

Now, we compute β for all signatures which satisfy 85
42
≤ A(d) ≤ 32

15
.

d = β 6

(2, 3,n) with 7 6 n 6 30 41
30

(2, 4,n) with 5 6 n 6 8 5
4

(2, 5,n) with 5 6 n 6 6 13
10

(3, 3,n) with 4 6 n 6 5 11
12

In the above table the maximum β is 41
30

. Now set β ≤ 41
30

and q ≥ 16. We
substitute them in inequality (4) and we obtain that A(d) ≤ 9064

4335
. From this,

we find all signatures d, which are the following:

d = β 6

(2, 3,n) with 7 6 n 6 13 5
4

(2, 4,n) with 4 6 n 6 6 5
4

(3, 3, 4) 11
12

Again, we choose the maximum β in the above table which is β ≤ 5
4
. So

we put β ≤ 5
4

and q ≥ 16 in inequality (4) and hence A(d) ≤ 3012
1445

. Therefore,
all signatures are (2, 3,d3) with 7 ≤ d3 ≤ 12, (2, 4, 5),(2, 4, 6), (3, 3, 4).

Finally, for each signature d we can compute β and A(d) and put in in-
equality (4). So we can solve it and obtaining the values of q.



genus zero of projective symplectic groups 201

Theorem 3.1. If PGL(4,q) or PSL(4,q) possesses genus zero system,
then one of the following holds:

(i) q ≤ 13;
(ii) d and q as shown in the following table

d β A(d) q

(2,3,7) 6/7 85/42 16, 17, 19, 23, 25, 27, 29, 31, 32, 37

(2,3,8) 25/24 49/24 16, 17, 19, 23, 25

(2,3,9) 8/9 37/18 16, 17

(2,3,10) 7/6 31/15 16, 17

(2,3,12) 5/4 25/12 16

(2,4,5) 21/20 41/20 16, 17, 19

(2,4,6) 5/4 25/12 16

(3,3,4) 5/4 25/12 16

The next results are devoted to compute indices of elements of order 2, 3,
4 and 5 in PSL(4,q). Let ed be an element of order d in G.

Lemma 3.2. In G = PSL(4,q):

(i) If 2 - q, then f(e2) = 0, ind e2 = N2 or f(e2) = 2q + 2, ind e2 =
N−(2q+2)

2
.

(ii) If 2 | q, then f(e2) = q2 + q + 1, ind e2 =
N−(q2+q+1)

2
or f(e2) = q + 1,

ind e2 =
N−(q+1)

2
.

Proof. There are at most two conjugacy classes of involutions in G. For
each such class, we give a representative e2. Let Z be the center of SL(4,q).

(i) Suppose that q is even. Note that, since Z = {I}, we can identify G
with SL(4,q). Take the involution

e2 =




0 1 0 0

1 0 0 0

0 0 1 0

0 0 0 1


 ,

whose only eigenvalue is 1. The corresponding eigenspace is

E1 =
{

(v2,v2,v3,v4)
T : v2,v3,v4 ∈ GF(q)

}
.



202 h.m.m. salih, r.m. hussein

Since it has dimension 3, from Table 1 we achieve that e2 has q
2 + q + 1 fixed

points. Therefore, ind e2 =
N−(q2+q+1)

2
. As representative of the other class,

we can take

e2 =




0 1 0 0

1 0 0 0

0 0 0 1

0 0 1 0


 ,

whose only eigenvalue is 1. The associated eigenspace, which has dimension
2, is

E1 =
{

(v2,v2,v4,v4)
T : v2,v4 ∈ GF(q)

}
.

From Table 1, we obtain that e2 has q + 1 fixed points, whence ind e2 =
N−(q+1)

2
.

(ii) Suppose that q is odd. The matrix

e2 =



−1 0 0 0

0 −1 0 0
0 0 1 0

0 0 0 1




is a non central element of SL(4,q) : its projective image in G has order 2.
The eigenvalues of e2 are 1 and −1. The associated eigenspaces are

E1 =
{

(0, 0,v3,v4)
T : v3,v4 ∈ GF(q)

}
,

E−1 =
{

(v1,v2, 0, 0)
T : v1,v2 ∈ GF(q)}.

They both have dimension 2: from Table 1, we get that e2 has 2q + 2 fixed
points. Hence, ind e2 =

N−(2q+2)
2

.

(iii) Suppose that q ≡ 3 (mod 4) (so Z = {±I}). In this case, we have
another conjugacy class of involutions. Take

e2 =




0 −1 0 0
1 0 0 0

0 0 0 −1
0 0 1 0




and note that g2 = −I ∈ Z. Hence, the projective image of e2 in PSL(4,q)
has order 2. The characteristic polynomial of e2 is (x

2 + 1)2, which has no
root in GF(q). From Table 1, we deduce that ind e2 =

N
2

.



genus zero of projective symplectic groups 203

In G = PSL(4,q), if 3 | q − 5, then there are two conjugacy classes of
elements of order 3. Otherwise there are four conjugacy classes of elements of
order 3.

Lemma 3.3. In G = PSL(4,q):

(i) If q ≡ 2 (mod 3), then f(e3) = 0, q + 1, ind e3 = 2N3 ,
2
3
(N −q − 1).

(ii) If q ≡ 1 (mod 3), then f(e3) ∈ {2q + 2,q + 3,q2 + q + 2}, ind e3 ∈{
2
3
(N − 2q − 2), 2

3
(N −q − 3), 2

3
(N −q2 −q − 2)

}
.

(iii) If q ≡ 0 (mod 3), then f(e3) ∈ {q + 1,q2 + q + 1}, ind e3 ∈
{

2
3
(N−

q − 1), 2
3
(N −q2 −q − 1)

}
.

Proof. Suppose element e3 has prime order 3 in G. Then all powers of e3
except the identity have the same fixed points. Now ind e3 =

2
3
(q3 + q2 + q +

1 −f(e3)) and ind e3 is an integer.
(i) Since 3 divides (q3 + q2 + q + 1 −f(e3)), this gives f(e3) ∈ {0, 3,q +

1, 2q + 2}. Next, we will show that 2q + 2 and 3 can not exist. We check
only the first 2q + 2. Suppose that v is an eigenvector of ē3 then ē3v = λv for
some nonzero number in GF(q). So v = Iv = (ē3)

3v = λ3v. So λ3 = 1. But
3 - q − 1, there is no element of order 3 in GF(q), we obtain λ = 1. So all
eigenvector of ē3 belong to eigenvalue 1. Suppose that ē3 fixes 2q + 2 points,
ē3 has two 2-dimensional eigenspaces. Both of them belong to 1. We get ē3 is
the identity. This is a contradiction. In similar way, proving 3 can not exist.

(ii) Since 3 divides (q3 + q2 + q + 1 −f(e3)), then f(e3) ∈{1,q + 3, 2q +
2,q2 + q + 2}. Next we will show that 1 can not exist. Since 3|q − 1, then ē3
is conjugate to one of the following:

α 0 0 0

0 β 0 0

0 0 1 0

0 0 0 1


 ,



α 0 0 0

0 α 0 0

0 0 β 0

0 0 0 β


 ,



α 0 0 0

0 α 0 0

0 0 α 0

0 0 0 1


 ,



β 0 0 0

0 β 0 0

0 0 β 0

0 0 0 1




where α = β−1 is a fixed element of order 3. This implies that f(e3) ∈
{q + 3, 2q + 2,q2 + q + 2}. Therefore, ind e3 ∈ {23 (N − 2q − 2),

2
3
(N − q −

3), 2
3
(N −q2 −q − 2)}.

(iii) Since 3 divides (q3 +q2 +q+ 1−f(e3)), so f(e3) ∈{1,q+ 1,q2 +q+ 1}.
In similar way, proving 1 can not exist. So ind e3 ∈{23 (N −q−1),

2
3
(N −q2 −

q − 1)}.



204 h.m.m. salih, r.m. hussein

Lemma 3.4. In PSL(4,q):

(i) If q ≡ 1 (mod 4), then f(e4) = f(e24) = 0 and ind e4 =
3N
4

, or f(e4) =

q + 1, f(e24) = 2q + 2 and ind e4 =
3N−(4q+4)

4
.

(ii) If q ≡ 0 (mod 4), then f(e4) = q + 1, f(e24) = q
2 + q + 1 and ind e4 =

3N−(q2+3q+3)
4

, or f(e4) = 1, f(e
2
4) = q + 1 and ind e4 =

3N−(q+3)
4

.

(iii) If q ≡ 3 (mod 4), then f(e4) ∈
{

0, 2q + 2,q + 3,q2 + q + 2,q + 3
}

, f(e24)

∈{2q + 2︸ ︷︷ ︸
3-times

,q2 + q + 2︸ ︷︷ ︸
2-times

} and ind e4 ∈
{

3N−(6q+6)
4

,
3N−(2q+2)

4
,

3N−(4q+8)
4

,

3N−3(q2+q+2)
4

,
3N−(q2+3q+8)

4

}
.

Proof. The proof is similar as Lemma 3.3.

Lemma 3.5. In PSL(4,q):

(i) If q ≡ 1 (mod 5), then f(e5) ∈
{
q + 3, 2q + 2,q2 + q + 2

}
and ind e5 =

4N−4f(e5)
5

.

(ii) If q ≡ 2 (mod 5) or q ≡ 3 (mod 5), then f(e5) = 0 and ind e5 = 4N5 .

(iii) If q ≡ 4 (mod 5), then f(e5) ∈{0,q + 1} and ind e5 = 4N5 ,
4N−4(q+1)

5
.

(iv) If q ≡ 0 (mod 5), than f(e5) ∈
{

1,q + 1,q2 + q + 1
}

and ind e5 =
4N−4f(e5)

5
.

Proof. The proof is similar as Lemma 3.3.

Table 2: Indices of some elements in PSL(4,q)

q 16 17 19 23 25

ind e6
3498, 3578,
3588, 3626

ind e8
4536, 4548,

4556, 4560, 4552
6324

11130,
11106, 11118

14222,
14214, 14226

ind e9 3822 4640, 4624, 4636

ind e10
3884, 3916,
3788, 3896

ind e12 3930



genus zero of projective symplectic groups 205

Proposition 3.6. In PSL(4,q), there is no generating tuple of genus zero
if 16 ≤ q ≤ 37.

Proof. From Theorem 3.1, we have to deal with seven possible signatures in
the different groups PSL(4,q). Since 7 - |PSL(4,q)| where q = 17, 19, 31, there
is no signature (2, 3, 7) in PSL(4,q) and 8 - |PSL(4, 16)|, there is no signature
(2, 3, 8) in PSL(4, 16). Also, 10 - |PSL(4, 17)|, there is no signature (2, 3, 10) in
PSL(4, 17). If q = 16, 23, 25, 32, 37, then f(e7) = 1 and ind e7 =

6N−6
7

. If q =

27, then f(e7) = 0, q + 1 and ind e7 =
6N
7
,

6N−6(q+1)
7

. If q = 29, then f(e7) =

2q + 2, q + 3, q2 + q + 2 and ind e7 =
6N−6(2q+2)

7
,

6N−6(q+3)
7

,
6N−6(q2+q+2)

7
.

We can compute the indices of elements of order 2 and 3 by Lemma 3.2 and
Lemma 3.3. The sum of the indices of the signature (2, 3, 7) does not fit the
Riemann Huwrtiz formula. By using Lemma 3.2, Lemma 3.3, Lemma 3.4 and
Lemma 3.5, we can compute the indices of elements of orders 2, 3, 4 and 5.
On the other hand, from Table 2, we can get the indices of the elements of
the other orders. Therefore, the sum of the indices of the given signatures do
not fit the Riemann Hurwitz formula. This completes the proof.

Lemma 3.7. The groups PSL(4,q) do not possess genus zero system,
if q = 7, 9.

Proof. The corresponding tuples of the following signatures satisfy the
Riemann Hurwitz formula (2, 3,d), d ∈ {7, 8, 9, 14, 16, 19, 24, 28, 42} and
(3, 3,d), d ∈{7, 8, 9, 14}. However none of them generate the group PSL(4, 7)
that is, do not satisfy (3). Also, the associated tuple of the signature (2, 4, 6)
fits the Riemann Hurwitz formula. It is not satisfied (3).

The following GAP codes can be used to show that there is no tuples
satisfying the Riemann Hurwitz formula:

cc:=List(ConjugacyClasses(group),Representative);;

N:=DegreeAction(group);;

ind:=List(cc,x->N-Length(Orbits(Group(x),[1..N])));;

ss:=Elements(ind);;

s:=Difference(ss,[ss[1]]);;

poss:=RestrictedPartitions(2N-2,s);

Lemma 3.8. The groups PSL(4,q) do not possess genus zero system
if q = 8, 11, 13.



206 h.m.m. salih, r.m. hussein

Proof. The proof is a straightforward computation.

Theorem 3.9. If G is the projective symplectic group with PSp(4,q) ≤
G ≤ Aut(PSp(4,q)), q > 5, then G does not possess genus zero system.

Proof. Since PSp(4,q) is a subgroup of PSL(4,q), then from Theorem 3.1,
Proposition 3.6, Lemma 3.7 and Lemma 3.8, we show that the group PSL(4,q)
does not possess genus zero system. So is PSp(4,q), as desired.

4. Connected components of the Hurwitz space

The details of the following can be found in [6]. The computation shows
that there are exactly 165 braid orbits of G. The degree and the number of the
branch points are given in Table 3. Furthermore, we discuss the connectedness
of the Hurwitz space for these groups.

Our main result is Theorem 4.1, which gives the complete classification of
primitive genus 0 systems of G.

Theorem 4.1. Up to isomorphism, there exist exactly 5 primitive genus
zero groups G with PSp(4,q) ≤ G ≤ Aut(PSp(4,q)) for q ≤ 5. The
corresponding primitive genus zero groups are enumerated in Tables 5,
6 and 4.

This will be done by both the proof in algebraic topology and calculations
of GAP (Groups, Algorithms, Programming) software [2].

Proposition 4.2. If G = PSp(4, 4).2 and |Ω| = 85A, then Hinr (G,C) is
connected.

Proof. Since we have just one braid orbit for all types C and the Nielsen
classes N(C) are the disjoint union of braid orbits. From [6, Proposition 2.4],
we obtain that the Hurwitz space Hinr (G,C) is connected.

The proof of the following proposition is similar as Proposition 4.2.

Proposition 4.3. If G = PSp(4, 3) is the projective symplectic group
and r > 3, then Hinr (G,C) is connected.

Proposition 4.4. If G = PSp(4, 4) and |Ω| = 120A, then Hinr (G,C) is
disconnected.



genus zero of projective symplectic groups 207

Proof. Since we have more than one braid orbits for some types C and
the Nielsen classes N(C) are the disjoint union of braid orbit. We obtain from
[6, Proposition 2.4] that the Hurwitz space Hinr (G,C) is disconnected.

The proof of the following proposition is similar as Proposition 4.4.

Proposition 4.5. If G = PSp(4, 5) and |Ω| = 156, then Hinr (G,C) is
disconnected.

Acknowledgements

The authors would like thank the referees, whose comments and sug-
gestions helped to improve the manuscript.

References

[1] J.D. Dixon, B. Mortimer, “ Permutation Groups ”, Graduate Text in
Mathematics, 163, Springer-Verlag, New York, 1996.

[2] The GAP Group, GAP – Groups, Algorithms, and Programming, Version
4.9.3, 2018. http://www.gap-system.org

[3] R.M. Guralnick, J. Thompson, Finite groups of genus zero, J. Algebra
131 (1) (1990), 303 – 341.

[4] X. Kong, Genus 0, 1, 2 actions of some almost simple groups of lie rank 2,
PhD Thesis, Wayne State University, 2011.

[5] K. Magaard, Monodromy and sporadic groups, Comm. Algebra 21 (12)
(1993), 4271 – 4297.

[6] H.M. Mohammed Salih, Hurwitz components of groups with socle
PSL (3,q), Extracta Math. 36 (1) (2021), 51 – 62.

[7] L.L. Scott, Matrices and cohomology, Ann. of Math. (2) 105 (3) (1977),
473 – 492.

http://www.gap-system.org


208 h.m.m. salih, r.m. hussein

Appendix

Table 3: Genus Zero Groups: Number of Components

Number of connected components

Degree
Number of

Group up to
Isomorphism

Number of
Ramification

Types
with r = 3 with r = 4 with r = 5 total

27 2 49 7 15 1 23

36 2 18 19 5 - 24

40A 2 20 39 4 - 43

40B 2 15 26 2 - 28

45 2 11 18 2 - 20

120A 1 1 4 - - 4

85A 1 1 1 - - 1

156A 1 2 22 - - 22

Total 13 117 136 28 1 165

Table 4: Genus Zero Systems for Projective symplectic Groups

Degree group
ramification

type N.O L.O
ramification

type N.O L.O

120A PSp(4, 4) (2A,4B,5E) 4 1

85A PSp(4, 4).2 (2C,4B,15A) 1 1

156A PSp(4, 5) (2B,4B,5B) 11 1 (2B,4B,5A) 6 1



genus zero of projective symplectic groups 209

Table 5: Genus Zero Systems for PSp(4, 3)

Degree ramification type N.O L.O ramification type N.O L.O

(2A,5A,6B) 1 1 (2A,5A,6A) 1 1

(2A,6F,9B) 3 1 (2A,6F,9A) 3 1

(2A,6F,12B) 1 1 (2A,6F,12A) 1 1

(2A,6D,9B) 1 1 (2A,6D,12B) 1 1

27 (2A,6C,9A) 1 1 (2A,6C,12A) 1 1

(2A,4B,9B) 1 1 (2A,4B,9A) 1 1

(2A,4A,9B) 3 1 (2A,4A,9A) 3 1

(2A,4A,12B) 3 1 (2A,4A,12A) 3 1

(2A,2A,2A,6B) 1 9 (2A,2A,2A,6A) 1 9

(2B,4B,9B) 3 1 (2B,4B,9A) 3 1

(2B,4B,12B) 3 1 (2B,4B,12A) 3 1

36 (2B,6B,5A) 1 1 (2B,6B,9B) 1 1

(2B,4A,9B) 1 1 (2B,4A,9A) 1 1

(2B,2B,2B,6B) 1 9 (2B,2B,2B,6A) 1 9

(2B,6B,5A) 1 1 (2B,6B,9B) 1 1

40A (2B,6A,5A) 1 1 (2B,6A,9B) 1 1

(2B,4A,5A) 1 1 (2B,4A,9B) 1 1

(2B,4A,9B) 1 1 (2B,4A,9A) 1 1

(2B,6C,5A) 1 1 (2B,6B,5A) 1 1
40B

(2B,5A,12B) 1 1 (2B,5A,12A) 1 1

(2B,5A,9B) 1 1 (2B,5A,9A) 1 1

(2B,4A,9A) 1 1 (2B,4A,9B) 1 1
45

(2B,6B,5A) 1 1 (2B,6A,5A) 1 1



210 h.m.m. salih, r.m. hussein

Table 6: Genus Zero Systems for PSp(4, 3) : 2

Degree ramification type N.O L.O ramification type N.O L.O

(2C,4A,10A) 1 1 (2C,5A,6A) 2 1
(2A,12B,9A) 1 1 (2D,6A,8A) 2 1
(2D,6A,10A) 2 1 (2C,4D,12A) 2 1
(2C,6G,12B) 2 1 (2C,4D,9A) 3 1
(2C,6G,10A) 2 1 (2C,6F,12A) 2 1
(2C,6F,9A) 3 1 (2C,4C,12B) 4 1
(2C,4C,8A) 6 1 (2C,4C,10A) 5 1

27 (2C,6E,12B) 3 1 (2C,6E,8A) 4 1
(2C,6E,10A) 4 1

(2D,2D,2C,6A) 1 24 (2D,2C,2C,6G) 1 48
(2D,2C,2C,4C) 1 112 (2D,2C,2C,6E) 1 78
(2D,2C,2C,4A) 1 20 (2A,2A,4A,6B) 1 1
(2C,2C,2C,4D) 1 96 (2C,2C,2C,6F) 1 108

(2A,2D,2C,12A) 1 12 (2A,2D,2C,9A) 1 18
(2A,2C,2C,12B) 1 24 (2A,2C,2C,8A) 1 32
(2A,2C,2C,10A) 1 30 (2A,2D,2C,2C,2C) 1 648

(2C,6F,5A) 2 1 (2C,4B,10A) 5 1
(2C,4B,8A) 6 1 (2C,4B,12B) 4 1

36 (2C,4A,10A) 1 1 (2B,6F,10A) 2 1
(2B,6F,8A) 2 1 (2C,2D,2D,4C) 1 112

(2C,2D,2D,4A) 1 20 (2C,2C,2D,6E) 1 24

(2D,4D,8A) 6 1 (2D,6E,12A) 1 1
(2D,6E,10A) 1 1 (2D,6D,8A) 2 1
(2D,6A,9A) 6 1 (2D,6A,5A) 4 1

40A (2D,6A,5A) 7 1 (2D,4B,9A) 3 1
(2D,4B,12B) 2 1 (2D,4A,10A) 1 1

(2C,2C,2D,6A) 1 24 (2C,2C,2D,4A) 1 20
(2C,2C,2C,4B) 1 96 (2C,2C,2C,6B) 1 234

(2D,4C,8A) 6 1 (2D,6E,5A) 7 1
(2D,6D,8A) 2 1 (2D,6C,5A) 2 1

40B
(2D,4A,10A) 1 1

(2D,2D,2C,4A) 1 20 (2A,2D,2D,5A) 1 35

(2B,4D,8A) 6 1 (2B,4B,10A) 1 1
(2B,4A,9A) 3 1 (2B,4A,12A) 2 1

45
(2B,6D,5A) 2 1

(2B,2B,2D,4B) 1 20 (2B,2B,2B,4A) 1 96


	Introduction
	Preliminary
	Existence of genus zero system
	Connected components of the Hurwitz space