ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 22 (2022) #P1.09 / 135–147
https://doi.org/10.26493/1855-3974.2257.6de
(Also available at http://amc-journal.eu)
Maximal order group actions on Riemann
surfaces
Jay Zimmerman * , Coy L. May
Towson University, 7800 York Road, Towson, USA
Received 20 February 2020, accepted 2 July 2021, published online 13 May 2022
Abstract
A natural problem is to determine, for each value of the integer g ≥ 2, the largest order
of a group that acts on a Riemann surface of genus g. Let N(g) (respectively M(g)) be the
largest order of a group of automorphisms of a Riemann surface of genus g ≥ 2 preserving
the orientation (respectively possibly reversing the orientation) of the surface.
The basic inequalities comparing N(g) and M(g) are N(g) ≤ M(g) ≤ 2N(g). There
are well-known families of extended Hurwitz groups that provide an infinite number of
integers g satisfying M(g) = 2N(g). It is also easy to see that there are solvable groups
which provide an infinite number of such examples.
We prove that, perhaps surprisingly, there are an infinite number of integers g such that
N(g) = M(g). Specifically, if p is a prime satisfying p ≡ 1 (mod 6) and g = 3p + 1 or
g = 2p+1, there is a group of order 24(g− 1) that acts on a surface of genus g preserving
the orientation of the surface. For all such values of g larger than a fixed constant, there are
no groups with order larger than 24(g − 1) that act on a surface of genus g.
Keywords: Riemann surface, genus, group action, NEC group, strong symmetric genus.
Math. Subj. Class. (2020): 57M60, 20F38, 20H10
1 Introduction
A finite group G can be represented as a group of automorphisms of a compact Riemann
surface. In most of the classical work, the group actions were required to preserve the
orientation of the Riemann surface. It is also possible to allow the group actions to reverse
the orientation of the surfaces.
Among the most interesting group actions for a particular value of the genus g are those
such that the orders of the groups are “large” relative to the genus g. A natural problem,
*Corresponding author.
E-mail addresses: jzimmerman@towson.edu (Jay Zimmerman), cmay@towson.edu (Coy L. May)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
136 Ars Math. Contemp. 22 (2022) #P1.09 / 135–147
then, is to determine, for each value of the integer g ≥ 2, the largest order of a group that
acts on a Riemann surface of genus g.
Let N(g) (respectively M(g)) be the largest order of a group of automorphisms of a
Riemann surface of genus g ≥ 2 preserving the orientation (respectively possibly reversing
the orientation) of the surface. Now suppose the group G acts on the Riemann surface X
of genus g ≥ 2 (possibly reversing the orientation of X). Let G+ be the subgroup of G
consisting of the orientation preserving automorphisms. Then |G+| ≤ N(g) and
|G| ≤ 2|G+| ≤ 2N(g). (1.1)
Consequently, we obtain the basic inequalities comparing N(g) and M(g).
N(g) ≤ M(g) ≤ 2N(g). (1.2)
The classical upper bound of Hurwitz shows that, for all g ≥ 2,
N(g) ≤ 84(g − 1) and M(g) ≤ 168(g − 1). (1.3)
A group G of order 84(g − 1) is called a Hurwitz group if it acts on a surface of genus g
preserving orientation. If the Hurwitz group has an extension G∗ of order 2|G| that acts on
the same surface, then G∗ is an extended Hurwitz group. If g is a genus for which there is
an extended Hurwitz group, then N(g) = 84(g − 1) and M(g) = 2N(g). These groups
have generated considerable interest; see especially [5] but also [3, 4] and [22]. There are
known infinite families of extended Hurwitz groups. For example, Conder showed that all
symmetric groups Σn for n > 167 are extended Hurwitz groups [8, p. 75]. Consequently,
the bounds in (1.3) and the upper bound for M(g) in (1.2) are attained for infinitely many
g.
On the other hand, the general lower bound for N(g) is
N(g) ≥ 8(g + 1) (1.4)
for all g ≥ 2. Further, this lower bound is the best possible, that is, there are infinitely many
g such that N(g) = 8(g + 1). These results were established independently by Accola [1]
and Maclachlan [13].
The corresponding lower bound for M(g) is easy to establish.
Theorem 1.1. For all integers g ≥ 2, M(g) ≥ 16(g + 1). Further, there are infinitely
many g such that M(g) = 16(g + 1).
The family of groups used by Accola [1] and Maclachlan [13] to establish the re-
sults about the lower bound (1.4) can be extended following the approach of Singerman
[22, p. 22]. This yields, for each g ≥ 2, the construction of a group of order 16(g+ 1) that
acts on a Riemann surface of genus g so that M(g) ≥ 16(g + 1).
These groups are another family of groups such that M(g) = 2N(g) for infinitely
many genera g. Indeed, intuitively, one expects M(g) to “often” be equal to 2N(g). But
it is certainly possible that M(g) < 2N(g). For example, the two smallest values of
g for which M(g) < 2N(g) are 17 and 20 with N(17) = 1344, M(17) = 1536 and
N(20) = 228, M(20) = 336. Also, the classification of orientably regular maps of genus
p+ 1 [9] and the Belolipetsky-Jones group of order 12p for prime p [2, p. 382] shows that
M(g) < 2N(g) for infinitely g.
J. Zimmerman and C. L. May: Maximal order group actions on Riemann surfaces 137
However, for some values of g, N(g) = M(g). The two smallest values of g satisfying
N(g) = M(g) are 27 and 28 with N(27) = M(27) = 624 and N(28) = M(28) = 1296.
Surprisingly, this equality holds for infinitely many g. Our main result is the following.
Theorem 1.2. There are infinitely many g such that M(g) = N(g).
Specifically, if p is a prime satisfying p ≡ 1 (mod 6) and g = 3p+ 1, there is a group
of order 24(g−1) that acts on a surface of genus g preserving the orientation of the surface.
For all such values of g = 3p + 1 larger than a fixed constant, there are no groups with
order larger than 24(g − 1) that act on a surface of genus g (including those that reverse
orientation). Similar results hold if p ≡ 1 (mod 6) and g = 2p+ 1.
Here we acknowledge our debt to the data on large group actions on surfaces of low
genus calculated by Conder [6]. This data was quite helpful in conjecturing Theorem 5.6
and its corollary Theorem 1.2.
We would also like to express our sincere gratitude to the referee for numerous helpful
comments. These led to significant improvements in the first three sections.
2 Background results
Much of the following background information is taken from [18]; also see
[10, Section 2]. Let the finite group G act on the (compact) Riemann surface X of genus
g ≥ 2. Then represent X = U/K, where K is a Fuchsian surface group and obtain
an NEC group Γ and a homomorphism ϕ : Γ → G onto G such that K = kernel ϕ.
Associated with the NEC group Γ are its signature and canonical presentation.
Further, the non-euclidean area µ(Γ) of a fundamental region for Γ can be calculated
directly from its signature. Here see [21, p.235], where µ(Γ) is given in terms of the
topological genus of the quotient surface U/Γ and the periods and link periods of Γ. Then
the genus of the surface X on which G acts is given by
g = 1 + |G| · µ(Γ)/4π. (2.1)
The simpler, classical case is that G acts on X preserving orientation. This is the case
if and only if Γ is a Fuchsian group and G is generated by elements ai, bi for 1 ≤ i ≤ h
and xj of order mj for 1 ≤ j ≤ k with relation x1 · · ·xk[a1, b1] · · · [ah, bh] = 1. Then the
application of (2.1) yields the classical Riemann-Hurwitz equation
2g − 2 = |G|
2h− 2 + k∑
j=1
(
1− 1
mj
) . (2.2)
The group G acts reversing the orientation of X in case Γ is a proper NEC group. Then
it is necessary to check that the surface group K does not contain orientation-reversing
elements, or equivalently, the image ϕ(Γ+) has index two in G [20, Theorem 1, p. 52].
If this condition holds, then we will say that G has a particular partial presentation with
the Singerman subgroup condition. The Riemann-Hurwitz equation in this case is more
complicated and is in [10, p. 274], for instance. In this case, though, |G| = 2|G+| and
(2.2) can be employed to calculate the relationship between the genus g and |G|.
In connection with group actions on surfaces, there are two natural parameters associ-
ated with each finite group. The symmetric genus σ(G) of the group G is the minimum
138 Ars Math. Contemp. 22 (2022) #P1.09 / 135–147
genus of any Riemann surface on which G acts faithfully (possibly reversing orientation).
The strong symmetric genus σ0(G) of G is the minimum genus of any Riemann surface on
which G acts faithfully preserving orientation.
Next we quickly survey the NEC groups with relatively small non-euclidean area. We
use the notation of [18]. First, an (ℓ,m, n) triangle group is a Fuchsian group Λ with
signature
(0;+; [ℓ,m, n]; {}), where 1/ℓ+ 1/m+ 1/n < 1.
If the group G is a quotient of Λ by a surface group, then G has a presentation of the form
Xℓ = Y m = (XY )n = 1. (2.3)
We will say that G has partial presentation T (ℓ,m, n).
There are two types of NEC groups with a triangle group as canonical Fuchsian sub-
group. A full (or extended) (ℓ,m, n) triangle group is an NEC group Γ with signature
(0;+; [ ]; {(ℓ,m, n)}), where 1/ℓ+ 1/m+ 1/n < 1.
If G is a quotient of Γ (by a surface group), then G has a presentation of the form
A2 = B2 = C2 = (AB)ℓ = (BC)m = (CA)n = 1, (2.4)
and, further, the subgroup generated by AB and BC (the image of Γ+) has index 2. The
partial presentation (2.4) will be denoted FT (ℓ,m, n).
A hybrid (m;n) triangle group is an NEC group Γ with signature
(0;+; [m]; {(n)}), where 2/m+ 1/n < 1.
The canonical Fuchsian subgroup Γ+ is a (m,m, n) triangle group. If G is a quotient of Γ,
then G has a presentation of the form
C2 = Xm = [C,X]n = 1, (2.5)
and the subgroup generated by X and CXC has index 2. This partial presentation will be
denoted HT (m;n).
An (ℓ,m, n, t) quadrilateral group is a Fuchsian group Λ with signature
(0;+; [ℓ,m, n, t]; { }), where 1/ℓ+ 1/m+ 1/n+ 1/t < 2.
A quotient group G of Λ has a presentation of the form
Xℓ = Y m = Zn = (XY Z)t = 1 (2.6)
We will denote this partial presentation Q(ℓ,m, n, t). If a group has presentation (2.6) with
ℓ, m, n, t all equal to 2, then the group acts on a torus.
Suppose G is a group that acts on a Riemann surface X of genus g ≥ 2, where X is
represented X = U/K and G = Γ/K. Particularly important here is the case in which
|G| > 24(g − 1), and we will say G is a large group of automorphisms of X . There
is, of course, a corresponding restriction on the non-euclidean area of the NEC group Γ
and the types of partial presentations that Γ can have. The area restriction is µ(Γ)/2π <
1/12, which is fairly limiting. A careful check of the signatures gives the following. This
result appears in [18, Theorem 2] and also [10, p. 275]. Here we have added the specific
Riemann-Hurwitz equation for each case. For example, if G has the partial presentation
FT (2, 4, s), then µ(Γ)/2π = (s − 4)/8s. Then using equation (2.2) gives 16(g − 1) =
|G|(s− 4)/s.
J. Zimmerman and C. L. May: Maximal order group actions on Riemann surfaces 139
Theorem A. Let G be a group that acts on a Riemann surface of genus g ≥ 2. Then
|G| > 24(g − 1) if and only if G has a partial presentation (with the relations fulfilled)
of type T (2, 4, 5) or T (2, 3, s), where 7 ≤ s ≤ 11, or one of the following types with the
Singerman subgroup condition satisfied. The application of the Riemann-Hurwitz equation
is included for each case.
1. FT (2, 3, s), 24(g − 1) = |G|(s− 6)/s where s ≥ 7,
2. FT (2, 4, s), 16(g − 1) = |G|(s− 4)/s where 5 ≤ s ≤ 11,
3. FT (2, 5, s), 20(g − 1) = |G|(3s− 10)/s where 5 ≤ s ≤ 7,
4. FT (3, 3, s), 12(g − 1) = |G|(s− 3)/s where 4 ≤ s ≤ 5,
5. HT (3; 4), 48(g − 1) = |G|,
6. HT (3; 5), 30(g − 1) = |G|,
7. HT (5; 2), 40(g − 1) = |G|.
3 Basic lower bound for M(g)
We begin by constructing the family of groups that provides the lower bound in Theo-
rem 1.1.
Fix the integer m ≥ 3, and let Lm be the group defined by the presentation
x2 = y4 = z2m = xyz = 1, (z2)x = z−2. (3.1)
It is easy to see that Lm is an extension of the cyclic group Zm by the dihedral group D4 and
consequently |Lm| = 8m. Then the group Lm has partial presentation T (2, 4, 2m). Then
a calculation using (2.1) shows that Lm acts on a Riemann surface X of genus g = m− 1
preserving the orientation of the surface. The group Lm has order 8(g + 1). This fam-
ily of groups is certainly not new. The family Lm was used, independently, to estab-
lish the lower bound 8(g + 1) by both Accola and Maclachlan; here see [1, p. 400] and
[13, Theorem 4, p. 266]. The construction of this family also appears in [2, p. 384].
Next we construct an extension of the group Lm by Z2, following the approach in
[15, p. 128]. To Lm adjoin an element t of order 2 that transforms the elements of Lm
according to the automorphism
α(x) = x−1, α(y) = y−1. (3.2)
Then the extension L∗m has presentation
t2 = x2 = y4 = z2m = xyz = (tx)2 = (ty)2 = 1, (z2)x = z−2. (3.3)
The extension L∗m of Lm has order 2|Lm| and has partial presentation FT (2, 4, 2m). Thus
the group L∗m is a group of order 16(g + 1) that acts on the surface X of genus g. This
extended family was described by Singerman in [22, p. 24]. Now it is easy to prove
Theorem 1.1.
Proof. Fix g ≥ 2, and set m = g + 1. Then M(g) ≥ |L∗m| = 16(g + 1). Also, there are
infinitely many values of g such that N(g) = 8(g + 1); here see [1, Theorem 4, p. 407]
or [13, Theorem 5, p. 272]. Then for such a value of g, M(g) ≤ 2N(g) = 16(g + 1)
using the basic inequality (1.2). Hence there are infinitely many values of g such that
M(g) = 16(g + 1).
140 Ars Math. Contemp. 22 (2022) #P1.09 / 135–147
Before proving Theorem 1.2, we establish an interesting result about the family of
groups Lm. We have seen that Lm acts preserving orientation on a Riemann surface of
genus g = m− 1. In fact, this value is the strong symmetric genus of the group Lm.
First, we get rid of a redundant generator in the definition of Lm and obtain the presen-
tation
x2 = z2m = (zx)4 = 1, (z2)x = z−2. (3.4)
Theorem 3.1. σ0(Lm) = m− 1.
Proof. Let Lm have generators x and z and relations (3.4) and be generated by u and v.
Define N = ⟨z2, (zx)2⟩. The element (zx)2 is in the center of Lm and since conjugation
by x inverts z2, N is a normal subgroup of Lm. Since Lm/N ∼= Z2 × Z2, uN , vN and
uvN are the same as the cosets xN , zN and zxN in some order. All elements of Lm in
the set xN have order 2 and all elements in the set zxN have order 4. The elements of the
set zN are of the form zk or zk(zx)2 and have order 2m/d, where d = gcd(k, 2m).
Let u be an element from xN and v an element from zxN . The product is contained
in zN . So uv = (xz)2zt = ((xz)2z)t or uv = zt, where t is odd. Next, suppose the
product uv has order smaller than 2m. So gcd(t,m) = d > 2. Let M = ⟨(xz)2, zt⟩.
Since xzx = (xz)2z−1, M is a normal subgroup of Lm of order 4m/d. It follows that
⟨u, v⟩ = ⟨u, uv⟩ ⊆ ⟨u,M⟩ and since |⟨u,M⟩| = 8m/d ̸= 8m, the elements u and v do
not generate Lm. Therefore, the product of two generators, u of order 2 and v of order 4,
must have order 2m.
Hence Lm has presentation T (2, 4, 2m), and the corresponding triangle group is the
only one that maps faithfully onto Lm.
Suppose that Lm has partial presentation Q(2, 2, 2, 2) and acts on a torus. Let Lm be
generated by involutions s, t, u and v satisfying stuv = 1. Let N = ⟨(zx)2, z2⟩. All
elements of order 2 are either contained in the coset xN or are in the normal subgroup
V = ⟨(zx)2, zm⟩ of Lm. Since there are no elements of order 2 in the coset zxN , an even
number of s, t, u or v must be from xN . If none of the generators are from xN , then
⟨s, t, u, v⟩ ⊆ ⟨z,N⟩ ̸= Lm. If all four generators are in xN , then ⟨s, t, u, v⟩ ⊆ ⟨x,N⟩ ≠
Lm. Suppose that only two of the generators are from xN , say u and v. Then ⟨s, t, u, v⟩ ⊆
⟨x, V ⟩ ̸= Lm and again we get a contradiction. It also follows that σ0(Lm) ̸= 1 for all
m > 2.
Suppose that Lm has partial presentation Q(2, 2, 2, 3). Let Lm be generated by involu-
tions s, t, u and the element v of order 3. The element v must be contained in ⟨z2⟩ ⊆ N .
Since there are no elements of order 2 in the coset zxN and (sN)(tN)(uN) = (1N), we
can’t have one of the cosets be xN and another be zN , since then the third would be in
zxN . If one or more of s, t and u are in xN , then ⟨s, t, u, v⟩ ⊆ ⟨x,N⟩ ≠ Lm. If one
or more of s, t and u are in zN , then ⟨s, t, u, v⟩ ⊆ ⟨z,N⟩ ≠ Lm. So Lm does not have
presentation Γ(2, 2, 2, 3). No other Fuchsian group has small enough non-euclidean area
and the proof is complete.
Theorem 3.1 shows that there is at least one group with strong symmetric genus g for all
g, which is the main result of [16]. One interesting thing here is that the well-known groups
of Theorem 3.1 provide an alternate proof of [16, Theorem 1], which was established using
groups of the form Zk ×Dn.
Theorem 3.1 also has a consequence for the function that counts the number of groups
of each genus. Using direct products and dicyclic groups, it was shown that there are at
J. Zimmerman and C. L. May: Maximal order group actions on Riemann surfaces 141
least four groups of strong symmetric genus g for all g ≥ 0 [14, Theorem 1]. It is not hard
to see that the group Lg+1 is another group of genus g, and we have the following.
Theorem 3.2. If g is a non-negative integer, then there are at least 5 groups of strong
symmetric genus g.
We remark here that these families Lm and L∗m are groups that act on the torus. These
groups are in in Proulx classes (g) and (k) respectively; for the associated partial presenta-
tions, see [12, pp. 291,292]. The orientation preserving subgroup of the action of L∗m on
the torus is not Lm, even though Lm is the orientation preserving subgroup of the action of
L∗m on the surface of genus m − 1. Consequently, the two families Lm and L∗m are of no
help in filling the symmetric genus spectrum. The groups Zk ×Dn used in [16] to fill all
the gaps in the strong symmetric genus spectrum are also groups that act on the torus and
have symmetric genus one.
4 A family of 24(g − 1) automorphisms
Our main task here is to show there are infinitely many values of g such that M(g) = N(g).
This result was something of a surprise, to us at least, and it is not easy to prove.
We start with the construction of another family of groups. Let p be a prime satisfying
p ≡ 1 (mod 6) and m an integer satisfying m3 ≡ 1 (mod p) and not congruent to 1
(mod p). Define the groups Jp by the presentation
x3 = u3 = v2 = zp = (uv)4 = [x, u] = [x, v] = [z, u] = 1, (4.1)
zx = zm, zv = z−1.
It is easy to see that Jp is the semidirect product of the cyclic group Zp by the group
Z3 ×Σ4, namely Zp ×ϕ (Z3 ×Σ4) where ϕ is a homomorphism mapping x into z → zm,
u into z → z and v into z → z−1.
Theorem 4.1. The group Jp has partial presentation T (2, 3, 12) and hence acts on a sur-
face of genus 1 + 3p.
Proof. We will use the presentation (4.1). First, o(v) = 2, o(ux) = 3 and o(vux) = 12.
In addition, ⟨v, ux⟩ = ⟨x, u, v⟩ ∼= Z3 × Σ4.
Define r = vz and w = z−1ux. Clearly, o(r) = 2. It is easy to verify that z−kx =
xz−km. Therefore, w3 = (z−1x)3 = z−(m
2+m+1) = 1, since m2 +m+ 1 ≡ 0 (mod p).
It follows that o(w) = 3 and o(rw) = 12.
Next, we need to show that Jp = ⟨r, w⟩. First, [r, w] = z−(m+1)[v, u]. Next, we show
that [r, w]3 = z−3(m+1). If m ≡ −1 (mod p), then m3 ≡ −1 (mod p) and this is false.
Therefore, z ∈ ⟨[r, w]⟩ and it follows that ⟨r, w⟩ = ⟨z, x, u, v⟩ = Jp. Thus Jp has partial
presentation T (2, 3, 12).
It is not difficult to see that, in fact, σ0(Jp) = 1 + 3p. An obvious consequence of
Theorem 4.1 is the following.
Theorem 4.2. Let p be a prime such that p ≡ 1 (mod 6), and let g = 3p + 1. Then the
group Jp is a group of order 24(g − 1) that acts on a surface of genus g preserving the
orientation of the surface. Consequently, for any such g,
M(g) ≥ N(g) ≥ 24(g − 1). (4.2)
142 Ars Math. Contemp. 22 (2022) #P1.09 / 135–147
There are, of course, infinitely many such g. We will show that, for most of these values
of g, M(g) = N(g) = 24(g − 1), establishing Theorem 1.2.
5 Large groups of automorphisms
Assume p is a prime such that p ≡ 1 (mod 6), and let g = 3p + 1. Then Theorem 4.2
shows that there is a group of order 24(g − 1) that acts on a surface of genus g preserving
the orientation of the surface, and inequality (4.2) holds. The hard part of the proof of
Theorem 1.2 is to show that, for most of these values of g, there are no large groups of
automorphisms, that is, no groups with order larger than 24(g− 1). We use Theorem A. In
this section we do not assume that p ≡ 1 (mod 6). However, in the proof it is necessary to
assume that that the prime p is not small. This will enable us to apply the following useful
result of Accola [1, Lemma 5, p. 402].
Accola’s Lemma. Let G be a non-abelian image of the triangle group T (2, 3, λ) of order
µλ. Then λ ≤ µ2.
Let X be a Riemann surface of genus g, and suppose that G were a large group of auto-
morphisms of X . Then |G| > 24(g − 1) = 72p, and G has one of the partial presentations
in Theorem A. We show that, in fact, G cannot have any of these partial presentations.
While it is necessary to consider each presentation, we describe the overall outline of the
argument but omit some details. In addition, to apply Accola’s Lemma, it is necessary to
assume that the prime p is not small, and we assume that p > (36)2.
Lemma 5.1. If the prime p > (36)2, then p divides |G| but p2 does not.
Proof. Suppose first that G has any of the partial presentations in Theorem A except
FT (2, 3, s). In these cases, the Riemann-Hurwitz formulas in Theorem A give |G| in terms
of the parameter s, and for the values of s that can occur, |G| is a multiple of p but p2 does
not divide |G| (for large p). For example, suppose G has partial presentation FT (2, 5, s),
where s is 5, 6 or 7. Then if s is 5 or 6, then G is 120p or 90p, respectively, and p2 does
not divide |G| if p > 5. If s = 7, then |G| is not an integer.
Suppose now that G has partial presentation FT (2, 3, s) where s ≥ 7. In this case,
|G| = 72ps/(s − 6) so that 72ps = |G|(s − 6). First, for small s, 7 ≤ s ≤ 12, s ̸= 11,
|G| is a multiple of p but p2 does not divide |G| (for large p). For example, if s = 8,
|G| = 288p. If s = 11, |G| is not an integer.
Assume then that G has partial presentation FT (2, 3, s) where s > 12, the hard case.
Now by Euclid’s Lemma, either p divides |G| or p divides (s− 6).
Assume that p divides (s − 6) and write s − 6 = mp for some integer m ≥ 1. Now
s = mp + 6 > p > (36)2 (by assumption). But on the other hand, |G| = 72ps/mp =
72s/m. Then |G+| = 36s/m. The group of orientation preserving automorphisms G+
is a T (2, 3, s) group of order cs, where c = 36/m ≤ 36. Now by Accola’s Lemma,
p < s ≤ c2 ≤ (36)2, an obvious contradiction. Thus, if G is a FT (2, 3, s) group (and
p > (36)2), then p divides |G|.
Finally, we have |G|/p = 72s/(s − 6). With s > 12, s/(s − 6) < 2 so that
|G|/p = 72s/(s− 6) < 144. Hence, p2 does not divide |G| for large p.
Lemma 5.2. The Sylow p-subgroup Sp ∼= Zp of G is normal in G.
J. Zimmerman and C. L. May: Maximal order group actions on Riemann surfaces 143
Proof. A review of the calculations in the previous proof shows that in each case that is
arithmetically possible, |G| = cp for some constant c < p for large p. The constant c
depends on the presentation, of course. Now, obviously, |Sp| = p. Also, the number np of
Sylow p-subgroups of G is ≡ 1 (mod p) and is a divisor of |G|. Then np ≥ (p + 1) is
clearly not possible. Hence Sp is normal in G.
Now let Sp act on X with Y = X/Sp the quotient space, γ the genus of Y and
π : X → Y the quotient map.
Lemma 5.3. The quotient map π is unramified, and the quotient space Y = X/Sp has
genus γ = 4. Further, the quotient group Q = G/Sp is a large group of automorphisms
of Y .
Proof. Let τ be the number of branch points of π. Then the Riemann-Hurwitz formula
gives
2(g − 1)/p = 2(γ − 1) + τ(p− 1)/p. (5.1)
Then 2(g − 1) = 2p(γ − 1) + τ(p − 1) and we have g − 1 = 3p. Now τ(p − 1) =
6p− 2p(γ − 1) = 2p(4− γ). Since τ(p− 1) ≥ 0, 4 ≥ γ. If γ = 4, then τ = 0. Assume
γ < 4. Then p− 1 divides p(8− 2γ). Since p− 1 and p are relatively prime, p− 1 divides
(8 − 2γ) so that p − 1 ≤ 8 − 2γ ≤ 8. Now p ≤ 9 contradicting the assumption that p is
large. Thus γ = 4 and the number of branch points τ = 0, that is, the quotient map π is
unramified.
Now the quotient group Q acts on the surface Y of genus 4. Since G is a large group
of automorphisms of X , |G| > 24(g − 1) = 72p. Then |Q| = |G|/p > 72 = 24(4 − 1)
and Q is a large group of automorphisms of Y .
The large group actions on Riemann surfaces of genus 4 have been classified, and these
are presented in Table 1. These group actions were considered in determining the groups
of symmetric genus 4; here see [18, pp. 4089,4090] and [10, p. 285]. With a single
exception, these actions correspond to groups of reflexible regular maps. A description
of the connection between groups of regular maps and large groups of automorphisms of
Riemann surfaces is in [17, p. 24]. The regular maps of genus 4 were first classified by
Garbe [11, p. 53]. These maps also appear in [7, Table 1]. In Table 1, we give the group
number in the MAGMA small groups library. Map symbols are from [7].
Table 1: Large Group Actions on Surfaces of Genus 4.
Group Order Library Partial Map G/G′
Number Presentation Symbol
Σ3 × Σ4 144 183 FT (2, 3, 12) R4.1 (Z2)2
Z2 × Σ5 240 189 FT (2, 4, 5) R4.2 (Z2)2
Σ5 120 34 T (2, 4, 5) R4.2 Z2
144 186 FT(2,4,6) R4.3 (Z2)3
D4 ×D5 80 39 FT(2,4,10) R4.4 (Z2)3
Z2 ×A5 120 35 FT(2,5,5) R4.6 Z2
Σ5 120 34 HT (5; 2) Z2
The group G is an extension of Sp ∼= Zp by Q. Since |Q| is relatively prime to p, the
group G is a semidirect product, by the Schur-Zassenhaus Lemma.
144 Ars Math. Contemp. 22 (2022) #P1.09 / 135–147
Lemma 5.4. G ∼= Zp ×ϕ Q.
The following is important here. The proof is an exercise using the definition of semidi-
rect product.
Lemma 5.5. Let H be the semidirect product K ×θ Q, and let L = kernel(θ). Then L is
normal in the big group H .
Theorem 5.6. Let p be a prime such that p > (36)2. There are no large groups of auto-
morphisms that act on a surface of genus g = 3p+ 1.
Proof. For each of the possibilities for Q, we show that G cannot have the relevant partial
presentation.
First suppose there is a group G of order 144p with partial presentation FT (2, 3, 12).
In particular, G is generated by involutions. Then G ∼= Zp ×ϕ Q, where Q ∼= Σ3 × Σ4.
Let L = kernel(ϕ). Since ϕ : Q → Aut(Zp) ∼= Zp−1, Q/L is cyclic. It follows that Q′ ⊂
L ⊂ Q. Now a calculation shows that the commutator quotient group Q/Q′ ∼= (Z2)2.
Hence L must have index 1 or 2 in Q, and L is normal in G by Lemma 5.5. If L = Q, then
G ∼= Zp × Q. Then G is obviously not generated by involutions, since Zp is not. Hence
[Q : L] = 2 and the quotient group G/L has order 2p so that G/L is isomorphic to either
Z2p or the dihedral group Dp. Since Z2p is not generated by involutions, we must have
G/L ∼= Dp. But Dp is not a quotient of a FT (2, 3, 12) group (the product of reflections
in Dp has order p or 1). Thus there is no group of order 144p with partial presentation
FT (2, 3, 12).
Essentially the same proof (using the same notation) shows that there are no groups of
order 80p with presentation FT (2, 4, 10) and also none of order 144p with presentation
FT (2, 4, 6). The only difference in each of the cases is that the commutator quotient group
Q/Q′ ∼= (Z2)3. But it still follows that L has index 1 or 2 in Q.
The proof is very similar but even easier in case there were a group G of order 120p
with presentation FT (2, 5, 5). Then G ∼= Zp ×ϕ Q, where Q ∼= Z2 × A5. Now Q′ ∼= A5
so that either L = Q or L = Q′. Again, as in the previous cases, L has index 1 or 2 in Q,
and it follows in the same way that this case is not possible either.
Now suppose there were such a group G with order 120p with partial presentation
T (2, 4, 5). Then G is generated by two elements of orders 2 and 5. Then G ∼= Zp ×ϕ Q,
where Q ∼= Σ5. Then Q′ ∼= A5, and thus either L = Q or L = Q′. If L = Q, then
G ∼= Zp ×Q. Then G is obviously not a quotient of a T (2, 4, 5) group, since Zp is not (Zp
has no elements of order 2). Hence L = Q′ and the quotient group G/L has order 2p so
that G/L is isomorphic to either Z2p or the dihedral group Dp. Neither group is a quotient
of T (2, 4, 5) groups; neither group has an element of order 5. Thus there is no group of
order 120p with partial presentation T (2, 4, 5).
Further, there is no group G of order 240p with partial presentation FT (2, 4, 5). If
there were such a group G acting on a surface X of genus 3p + 1, then the group G+ of
orientation preserving automorphisms would be a T (2, 4, 5) group acting on X . But we
have just seen that this is not possible. Hence there is no FT (2, 4, 5) group of order 240p.
Finally, assume that there were a group G of order 120p with partial presentation
HT (5; 2). Then G is generated by two elements of orders 2 and 5, and G ∼= Zp ×ϕ Q,
where Q ∼= Σ5. Now Q′ ∼= A5, and either L = Q or L = Q′. If L = Q, then G ∼= Zp×Q.
Then G is obviously not a quotient of a HT (5; 2) group, since Zp is not (Zp has no ele-
ments of order 2). Hence L = Q′ and the quotient group G/L has order 2p so that G/L is
J. Zimmerman and C. L. May: Maximal order group actions on Riemann surfaces 145
isomorphic to either Z2p or the dihedral group Dp. Neither of these groups is a quotient of
a HT (5; 2) group; neither group has an element of order 5. Thus there is no group of order
120p with partial presentation HT (5; 2).
In summary, none of the partial presentations listed in Table 1 are possible, and there is
no large group action on a surface of genus g = 3p+ 1 for large p > (36)2.
Combining Theorems 4.2 and 5.6 gives the following.
Theorem 5.7. Let p be a prime such that p ≡ 1 (mod 6) and p > (36)2, and let
g = 3p+ 1. Then for any such g,
M(g) = N(g) = 24(g − 1). (5.2)
Applying Dirichlet’s Theorem about the number of primes in an arithmetic sequence
establishes Theorem 1.2.
6 Another family of 24(g − 1) automorphisms
There is another interesting family of groups that can be used to determine an infinite
sequence of odd values of g such that M(g) = N(g). This provides an alternate proof of
Theorem 1.2, and it may be established using arguments similar to those in the two previous
sections. But there is no improvement to Theorem 1.2, of course, and technically the proof
is somewhat harder. We describe this family of groups but only comment very briefly on
the arguments in this case.
Let p be a prime satisfying p ≡ 1 (mod 6) and m an integer satisfying m3 ≡ 1
(mod p) and not congruent to 1 (mod p). Define the groups Kp by the presentation
u3 = v2 = (uv)3(u−1v)3 = zp = 1, zu = zm, zv = z−1. (6.1)
It is easy to see that Kp is the semidirect product of the cyclic group Zp by the group
P48. The group P48 has order 48 and contains SL(2, 3) as a subgroup; a presentation is in
[15, p. 116]. It is one of the groups of symmetric genus 2 [15, Theorem 4]. The group Kp
has partial presentation T (2, 3, 12) and acts on a surface of genus 1 + 2p. This gives the
following analog of Theorem 4.2.
Theorem 6.1. Let p be a prime such that p ≡ 1 (mod 6), and let g = 2p + 1. Then the
group Kp is a group of order 24(g − 1) that acts on a surface of genus g preserving the
orientation of the surface. Consequently, for any such g,
M(g) ≥ N(g) ≥ 24(g − 1). (6.2)
Using the approach (and notation) of Section 6, it is possible to show that there are no
large groups of automorphisms for most of these values of g. The analog of Lemma 5.1
holds; it is necessary to assume that p > (24)2 to apply Accola’s result. Then the analog
of Lemma 5.2 is easy to establish. The result corresponding to Lemma 5.3 holds with a
similar proof. There is an important difference here, though. The quotient space Y has
genus γ = 3. It is necessary, then, to consider the large group actions on Riemann surfaces
of genus 3. These actions have been classified; see [18, p. 4089] and [10, p. 285]. The
regular maps of genus 3 were classified by Sherk [19]; also see [7, Table 1]. There are 10
large group actions in all. Eight of these are map groups, but there are also two groups of
96 to consider, a FT (3, 3, 4) group and a HT (3; 4) group. The analogs of Lemmas 5.4
and 5.5 continue to hold, as does the following companion to Theorem 5.6.
146 Ars Math. Contemp. 22 (2022) #P1.09 / 135–147
Theorem 6.2. Let p be a prime such that p > (24)2. There are no large groups of auto-
morphisms that act on a surface of genus g = 2p+ 1.
It is necessary to consider the ten possibilities for the quotient group Q. In nine of the
cases, as in the proof of Theorem 5.6, an argument using the commutator quotient group
suffices; in these cases, Q/Q′ is isomorphic to 1, Z2, (Z2)2, or (Z2)3. The exceptional
case is the one in which G is a group of order 96p with partial presentation HT (3; 4). In
this case, Q/Q′ ∼= Z6; this case can be handled by considering the group G+. Then it
is not hard to show that there is no T (3, 3, 4) group of order 48p and hence no HT (3; 4)
group of order 96p.
Combining Theorems 6.1 and 6.2 gives the following.
Theorem 6.3. Let p be a prime such that p ≡ 1 (mod 6) and p > (24)2, and let
g = 2p+ 1. Then for any such g,
M(g) = N(g) = 24(g − 1). (6.3)
Finally, it is worth noting that there are genera in which N(g) = M(g) but the genus g
does not have either the form 2p + 1 or the form 3p + 1 for a prime p. Two examples are
genus 28 and genus 37.
ORCID iDs
Jay Zimmerman https://orcid.org/0000-0003-2749-0057
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