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ARS MATHEMATICA CONTEMPORANEA 4 (2011) 271–289
The symmetric genus spectrum of finite groups
Marston D. E. Conder ∗
Mathematics Department, University of Auckland
Private Bag 92019, Auckland 1142, New Zealand
Thomas W. Tucker
Mathematics Department, Colgate University, Hamilton, NY 13346, USA
Received 10 December 2009, accepted 16 July 2011, published online 9 August 2011
Abstract
The symmetric genus of the finite group G, denoted by σ(G), is the smallest non-
negative integer g such that the group G acts faithfully on a closed orientable surface of
genus g (not necessarily preserving orientation). This paper investigates the question of
whether for every non-negative integer g, there exists some G with symmetric genus g. It
is shown that that the spectrum (range of values) of σ includes every non-negative integer
g 6≡ 8 or 14 mod 18, and moreover, if a gap occurs at some g ≡ 8 or 14 modulo 18, then the
prime-power factorization of g − 1 includes some factor pe ≡ 5 mod 6. In fact, evidence
suggests that this spectrum has no gaps at all.
Keywords: Symmetric genus, Riemann surface, Riemann-Hurwitz equation, NEC group, signature.
Math. Subj. Class.: 57M60; 14H37, 20B25, 20F38, 20H10
1 Introduction
The automorphisms of a compact Riemann surface S of genus g > 1 depend on g (and on
the analytic structure of S). A celebrated theorem of Hurwitz (1893) states that if G is any
group of automorphisms of such a surface preserving orientation, then |G| ≤ 84(g−1); see
[20, 18]. Similarly, Wiman’s theorem [19, 35] shows that |G| ≤ 2(2g+1) whenG is cyclic.
Both of these bounds are easy consequences of what is now known as the Riemann-Hurwitz
equation. They also stimulated work by Burnside and Maschke and others on the actions
of specific kinds of groups on surfaces, and on embeddings of graphical representations of
groups on surfaces; see [5, 26].
∗Supported in part by the N.Z. Marsden Fund (grant no. UOA 721).
E-mail addresses: m.conder@auckland.ac.nz (Marston D. E. Conder), ttucker@colgate.edu (Thomas W.
Tucker)
Copyright c© 2011 DMFA Slovenije
272 Ars Math. Contemp. 4 (2011) 271–289
Shortly afterwards, Brahana [4] initiated the study of regular maps on surfaces, which
generalize the Platonic solids. A map is a dissection of a surface into vertices, edges and
faces, and is called regular (or more technically, rotary) if it has symmetries that act as full
cyclic rotations about any face and any vertex.
These things are now much better understood than in the early 1900s. It is now known
exactly which kinds of groups arise as finite groups of homeomorphisms of the sphere
(genus 0) and the torus (genus 1); see [18]. Also much is known about the largest fi-
nite groups of automorphisms of a surface of given genus; see [1, 25, 9, 12] for example.
Significant progress has been made recently through the study of Fuchsian groups and non-
Euclidean crystallographic groups (see [23]) and their quotients, with the help of advanced
methods such as voltage graph constructions (see [18]) and computation (see [10] for ex-
ample). Further relevant comments will be made in the final section.
Given a finite group G, one may ask what is the smallest non-negative integer g for
which the group G acts faithfully on a closed orientable surface of genus g. The smallest
such g is now known as the symmetric genus of G, and denoted by σ(G); see [18, 32].
The definition allows G to contain both orientation-preserving and orientation-reversing
elements, and if we require that the action be orientation-preserving, then we have the
strong symmetric genus, denoted by σo(G). Similarly, for closed non-orientable surfaces
we have the symmetric cross-cap number, denoted by σ̃(G).
Determination of σo(G) and/or σ(G) for various candidates for G was a question of
interest to Burnside et al, and has been answered for several classes of finite groups, in-
cluding cyclic and dihedral groups (of strong symmetric genus 0), abelian groups [24, 27],
the alternating and symmetric groups [7, 8], other finite Coxeter groups [21], groups of odd
order [29], the projective special linear groups L2(q) [17], and the sporadic finite simple
groups (see [14, 34] and other references therein).
A further natural question to ask is this:
Question 1.1. For every integer g ≥ 0, is there is a finite group G with σ(G) = g? In
other words, for all g ≥ 0, is there some group G that acts faithfully on a closed orientable
surface of genus g, but on no such surface of genus smaller than g?
The analogous question for the strong symmetric genus σo(G) has been answered in
the affirmative by May and Zimmerman [28], who showed that σo(G) takes all possible
integer values g ≥ 0 when G is restricted to the family of direct products Cm × Dn (of
order 2mn). Subsequently, Etayo and Martinez [16] have shown that the same family
gives values of the symmetric cross-cap number σ̃ that cover a significant proportion of the
positive integers.
On the other hand, σ(Cm ×Dn) = 1 for all m and n, since these groups act faithfully
on the torus (with action of type (h) in the classification of toroidal groups given in [18]),
so these groups are of no use for answering Question 1.1 for σ.
We can, however, show that the spectrum (range of values) of the symmetric genus
contains well over 88% of all positive integers. Specifically, in this paper we prove the
following:
Theorem 1.2. If g is any non-negative integer such that g 6≡ 8 or 14 mod 18, then there
exists a finite group G with symmetric genus σ(G) = g. Moreover, the same holds if g ≡ 8
or 14 modulo 18 and every factor pe in the prime-power factorization of g−1 is congruent
to 1 mod 6.
M. D. E. Conder and T. W. Tucker: The symmetric genus spectrum of finite groups 273
Hence the only possible gaps in the spectrum for σ (if there are any gaps at all) must
occur at integers g ≡ 8 or 14 mod 18 such that g−1 has a factor pe ≡ 5 mod 6 in its prime-
power factorization. Evidence suggests, however, that there are no gaps. In fact we have
found none, and so we join May and Zimmerman [28] in making the following conjecture:
Conjecture 1.3. For every integer g ≥ 0, there is a finite group G with σ(G) = g.
The challenge in proving this conjecture is the difficulty of finding families of groups
whose orders grow in a well-behaved way and whose symmetric genera can be computed.
Knowing the symmetric genus of particular families of groups is helpful, but often only to
a certain extent. For the alternating and symmetric groups for example, it is known that
σo(An) = σ(An) = σ(Sn) = n!/168 + 1 for all but finitely many n (as follows from
[6]), but the orders of these groups (and apparently other Hurwitz groups [9]) grow expo-
nentially. We have found several good families of groups whose orders grow in arithmetic
progression, as we will demonstrate in the proof of Theorem 1.2, but sadly, there are not
quite enough of them.
The organization of this paper is as follows. In Section 2, we provide some further back-
ground on the Riemann-Hurwitz equation, signatures of group actions, and non-Euclidean
crystallographic groups. In Section 3 we show that σ(G) can be any odd positive integer.
Then in Section 4 we consider all even genera g 6≡ 8, 14 mod 18, while in Section 5 we
handle those g ≡ 8, 14 mod 18 for which g − 1 has no factor pe ≡ 5 mod 6 in its prime-
power factorization. Finally, in Section 6, we consider the remaining gaps in the spectrum
of σ, and other matters such as the genus spectra of various kinds of regular maps, and the
spectrum of the plain (Cayley) genus of a group, as defined by White [33].
2 Signatures and the Riemann-Hurwitz equation
Let G be a finite group of homeomorphisms of a closed orientable surface S. We con-
sider three possibilities: one where the action of G has a finite number of fixed points on
S, which happens when all elements of G preserve orientation; a second where the action
includes reflections (which are involutory orientation-reversing homeomorphisms that fix
at least one simple close curve in S) and the quotient surface S/G (obtained by identi-
fying each orbit of S under the action of G to a point) is orientable; and a third where
the action includes orientation-reversing homeomorphisms but the quotient surface S/G is
non-orientable.
In the first case, the Riemann-Hurwitz equation says
χ(S) = |G|(χ(S/G)−
r∑
i=1
(1− 1/mi)),
where χ denotes the Euler characteristic of the relevant surface, and m1, . . . ,mr are the
orders of the branch points in S/G. (Here the natural map S → S/G is a branched cov-
ering.) This equation is a simple consequence of counting vertices, faces, and edges of an
appropriate dissection of the surface S/G, taking into account that the ith branch point has
only |G|/mi preimages in S and hence has a deficiency of |G|(1− 1/mi), for 1 ≤ i ≤ r).
In terms of the genus, the equation says
g = 1 +
|G|
2
(2γ − 2 +
r∑
i=1
(1− 1/mi)),
274 Ars Math. Contemp. 4 (2011) 271–289
where g is the genus of S and γ is the genus of S/G (given by χ(S) = 2 − 2g and
χ(S/G) = 2− 2γ).
In this case, we say that the group G acts with signature (γ; +; [m1, ...,mr]; {−}).
Associated with this signature is a finitely-presented group Γ, called a Fuchsian group, of
the form
Γ = 〈 a1, b1, ..., aγ , bγ , x1, ..., xr | xm11 = ... = xmrr = [a1, b1]...[aγ , bγ ]x1...xr = 1 〉,
with the property that G is a quotient of Γ, under an epimorphism θ : Γ→ G that preserves
the orders mi of the generators xi. Such an epimorphism is often called smooth. From
the uniformization viewpoint, G ∼= Γ/Λ and S is the quotient space U/Λ of the upper
half-plane U by Λ = ker θ, which is the fundamental group of S.
In the second case, which has reflections, the Riemann-Hurwitz equation says
χ(S) = |G|(χ(S/G)−
r∑
i=1
(1− 1/mi)−
k∑
j=1
sj∑
`=1
(1− 1/nj`)/2),
where the quotient surface S/G has k (> 0) boundary components, the mi are the orders
of the r branch points in the interior of S/G, and the nj` are the orders of the sj branch
points on the jth boundary component Cj of S/G, for 1 ≤ j ≤ k. This is again a simple
consequence of counting vertices, faces, and edges, but here a branch point of order nj` on
the boundary has deficiency |G|(1− 1/nj`)/2. As before, the Riemann-Hurwitz equation
can also be re-expressed in terms of the genera of S and S/G, which are χ(S) = 2 − 2g
and χ(S/G) = 2− 2γ − k.
In this case, the signature is (γ; +; [m1, ...,mr]; {(n11, ..., n1s1), . . . , (nk1, ..., nksk)}),
and associated with it is a non-Euclidean crystallographic group (or NEC group) Γ, gener-
ated by elements a1, b1, ..., aγ , bγ , x1, ..., xr, e1, ..., ek, c10, ..., c1s1 , . . . , ck0, ..., cksk sub-
ject to defining relations
xmii = 1 for 1 ≤ i ≤ r, c2j` = 1 for 1 ≤ j ≤ k and 0 ≤ ` ≤ sj ,
ejcj0e
−1
j = cjsj for 1 ≤ j ≤ k, (cj(`−1)cj`)nj` = 1 for 1 ≤ j ≤ k and 1 ≤ ` ≤ sj ,
and [a1, b1]...[aγ , bγ ]x1...xr e1...ek = 1.
Again G is a (smooth) quotient of Γ, under an epimorphism θ : Γ → G that preserves
the orders of the generators and their products when these orders are specified, and S
is the quotient space U/Λ where Λ = ker θ is the fundamental group of S. This time,
however, there is another requirement: Γ has a unique subgroup Γo of index 2 that contains
all of a1, b1, ..., aγ , bγ , x1, ..., xr, e1, ..., ek but none of the involutory generators cj`, and
this subgroup is taken by the epimorphism θ to the orientation-preserving subgroup of G,
necessarily of index 2 in G.
The third case is similar to the second, but with the ‘+’ replaced by ‘−’ in the signature,
the generators a1, b1, ..., aγ , bγ of Γ replaced by generators d1, ..., dγ , and the final defining
relation for Γ replaced by
d1
2...dγ
2 x1...xr e1...ek = 1,
while Γo is taken as the unique subgroup of index 2 containing all of x1, ..., xr, e1, ..., ek
but none of d1, .., dγ and none of the involutory generators cj`. The Riemann-Hurwitz
equation is the same as in the second case, except that χ(S/G) = 2− γ − k.
M. D. E. Conder and T. W. Tucker: The symmetric genus spectrum of finite groups 275
These observations work in reverse, in the sense that if G is any smooth quotient of a
non-Euclidean crystallographic group Γ, then G acts faithfully on a closed surface whose
characteristic is given by the Riemann-Hurwitz equation.
Now in each of the theorems in the next few sections of this paper, we will exhibit a
faithful action of a particular finite group G on an orientable surface S′ of genus g′, and
then show G has no faithful action on any surface S of genus g smaller than g′.
In all cases, except where otherwise noted, we will have |G| ≥ 4(g′ − 1), and so to
improve on the genus g′, we would require g < g′ ≤ 1 + |G|/4. Linking this bound with
the Riemann-Hurwitz equation puts a strong restriction on the parameters involved in the
signatures of possible group actions, and hence also on the NEC-groups that can give rise
to such actions of G within the desired genus range.
Such restrictions are illustrated in the following proposition, which will be useful for
us later. Note that in each case the genus g is given by g = 1 + |G|ξ/2, for some rational
number ξ that depends on the signature of the corresponding action. If ξ is negative then
g = 0 and the group G acts on the sphere, while if ξ = 0 then g = 1 and the action is
toroidal.
Proposition 2.1. Let G be a finite group that is neither cyclic nor abelian of rank 2, nor
generated by involutions, and suppose G acts faithfully on an orientable surface of genus
g = 1 + |G|ξ/2 where ξ < 1/2. Then one of the cases below occurs :
(A) G acts with signature (0; +; [m1, . . . ,mr]; {−}), where r = 3 or 4,
and ξ = r − 2−
∑r
i=1 1/mi ;
(B1) G acts with signature (0; +; [m]; {(q1, . . . , qs)}) where s = 1, 2 or 3,
and ξ = s/2− 1/m−
∑s
j=1 1/(2qj) ;
(B2) G acts with signature (0; +; [m1,m2]; {(q)}) where m1 = 2 or q = 1,
and ξ = 3/2− 1/m1 − 1/m2 − 1/(2q) ;
(B3) G acts with signature (0; +; [−]; {(1), (q)}), and ξ = 1/2− 1/(2q) ;
(C1) G acts with signature (1;−; [m1,m2]; {−}), and ξ = 1− 1/m1 − 1/m2 ;
(C2) G acts with signature (1;−; [−]; {(q)}), and ξ = 1/2− 1/(2q) ;
(C3) G acts with signature (2;−; [−]; {−}), and ξ = 1 .
In cases (B1), (B3) and (C2), and in case (B2) with m1 = 2, the quotient G/J of G by
the normal subgroup J generated by involutions is cyclic, while in case (C3), and in cases
(B3) and (C2) with q = 1, the quotient G/Z(G) is cyclic or dihedral, and the group G has
a subgroup of index 2 that is abelian of rank at most 2.
Proof. First suppose the action of G preserves orientation. Then the signature is (γ; +;
[m1, ...,mr]; {−}), and we need 2γ − 2 +
∑r
i=1(1− 1/mi)) < 1/2, which forces γ = 0
or 1 (so S/G is the sphere or the torus). But also 1 − 1/mi ≥ 1/2 for all i, and hence we
find that if γ = 0 then r ≤ 4, while if γ = 1 then r = 0. If γ = 0 and r ≤ 1 then the
NEC-group Γ is trivial, while if γ = 0 and r = 2 then Γ is cyclic (generated by elements x1
and x2 such that x1x2 = 1), and this case can be ruled out since G is not cyclic. Similarly,
if γ = 1 and r = 0 then Γ is abelian of rank at most 2 (generated by elements a1 and b1
such that [a1, b1] = 1), so this case is ruled out too. Thus S/G is the sphere, with three or
four branch points, giving case (A).
276 Ars Math. Contemp. 4 (2011) 271–289
Next, suppose the action has reflections and S/G is orientable. Then the Riemann-
Hurwitz equation gives χ(S/G) = 2γ + k − 2 < 1/2 where k ≥ 1, so γ = 0 and
k = 1 or 2; in other words, S/G is the sphere, and there can be only one or two boundary
components. Moreover, if k = 1 then if s is the total number of branch points of order
greater than 1 on the boundary component, then 2r + s ≤ 5 (since (1 − 1/nj`)/2 ≥ 1/4
whenever nj` > 1). When r = 0 the NEC-group Γ is generated by involutions (which
induce reflections of S), and so we can rule out that case; when r = 1 we have case (B1);
and when r = 2 we have (B2), where we note that if m2 ≥ m1 ≥ 3 and q > 1 then
ξ = 3/2 − 1/m1 − 1/m2 − 1/(2q) ≥ 7/12 > 1/2, which is impossible. Similarly, if
k = 2 then r = 0 and s ≤ 1, giving (B3).
In case (B1), the NEC group Γ is generated by s involutions c0, . . . , cs−1 and an ele-
ment x such that xm = (c0c1)q1 = · · · = (cs−2cs−1)qs−1 = (cs−1x−1c0x)qs = 1. In
particular, the quotient of Γ by the subgroup generated by its involutions is cyclic (gener-
ated by the image of x), and so the analogous property holds in G. Similarly, in case (B2),
the NEC group Γ is generated by an involution c and two elements x1 and x2 such that
x1
m1 = x2
m2 = [c, x1x2]
q = 1, and so if q = 1 then the quotient of G by the subgroup
generated by its involutions is cyclic. In case (B3), the NEC group Γ is generated by two
involutions c1 and c2 and an element e such that [e, c1] = [e, c2]q = 1, so again the quotient
of G by the subgroup generated by its involutions is cyclic. Moreover, if q = 1 then e is
central and so G/Z(G) is cyclic or dihedral (generated by the images of c1 and c2), and
also the orientation-preserving subgroup of G is the image of the index 2 subgroup of Γ
generated by e and c1c2, which is abelian.
Finally, suppose S/G is non-orientable. Then the Riemann-Hurwitz equation gives
χ(S/G) = γ + k − 2 < 1/2, where γ ≥ 1. If γ = 1 and k = 0 then r ≤ 2, but if r = 0 or
1 then the NEC group Γ is cyclic, so we have r = 2, which gives case (C1). If γ = k = 1
then r = 0, and moreover, there can be at most one branch point of order greater than 1
on the boundary component, and so we have (C2). The only other possibility is γ = 2 and
k = 0, and then r = 0 and we have (C3).
In case (C2), the NEC group Γ (and analogously, its quotient G) is generated by an
involution c and an element d such that [c, d2]q = 1, and again, it follows that the quotient
of G by the subgroup generated by its involutions is cyclic. Moreover, if q = 1, then d2
is central, so G/Z(G) is cyclic or dihedral, and the index 2 subgroup 〈d2, cd〉 (preserving
orientation) is abelian. In case (C3), the NEC group Γ is generated by two elements d1 and
d2 such that d12d22 = 1. In particular, the element d12 = d2−2 is central, making G/Z(G)
cyclic or dihedral, and the images of d12 and d1d2 generate an abelian subgroup of index 2
in G (preserving orientation). 
3 Odd genera
In this section, we consider two families of groups defined as follows:
Un = 〈x, y | x4 = y4 = 1, [x2, y] = [y2, x] = 1, (xy)2n = 1 〉, and
Vn = 〈x, y | x4 = y4 = 1, [x2, y] = [y2, x] = 1, (xy)2n = x2 〉,
for each positive integer n.
Note that in each case, the subgroup N generated by x2 and y2 is central, with dihedral
quotient 〈x, y |x2 = y2 = (xy)2n = 1 〉 ∼= D2n, of order 4n, in which the image of xy has
order 2n. Similarly, the subgroup generated by (xy)2 is normal, because the centrality of
x2 and y2 implies that (xy)2 = (x−1y−1)2 and hence that (xy)2 is inverted by conjugation
M. D. E. Conder and T. W. Tucker: The symmetric genus spectrum of finite groups 277
by each of x and y. The quotient Un/〈(xy)2〉 is the group U1, which has a transitive
permutation representation of degree 8 given by x 7→ (1, 2, 3, 4)(5, 6, 7, 8) and y 7→
(1, 6, 5, 2)(3, 8, 7, 4), and this shows that in Un, the subgroup generated by x2 and y2 has
order 4 (and that U1 has order 16). Thus Un has order 16n. Moreover, Vn is isomorphic
to the quotient of U2n by its central subgroup of order 2 generated by (xy)2nx−2, and
therefore Vn has order 16n as well.
Next we make an important observation about the involutions in each of these groups.
Again since x2 and y2 are central, every element w in G can be expressed in the form
(xy)iz or (xy)ixz where 0 ≤ i < 2n and z ∈ {1, x2, y2, x2y2}, and then
w2 = ((xy)iz)2 = (xy)2i or ((xy)ixz)2 = x2i+2y2i.
In the first case, w is an involution only when i = 0 or n, while in the second case, w has
order 4 since w2 = x2 or y2, depending on whether i is even or odd (respectively). Hence
the only involutions in Vn are x2, y2 and x2y2, while the involutions in Un are the seven
non-trivial elements of the form z or (xy)n/2z with z ∈ {1, x2, y2, x2y2}.
Theorem 3.1. σ(Vn) = 4n− 1 for every integer n > 1.
Proof. Let G = Vn where n > 1, and let I be the set of elements of order 1 or 2 in G
(namely I = {1, x2, y2, x2y2}). The generating set {x, y, (xy)−1} gives an action of G
with signature (0; +; [4, 4, 4n]; {−}) on an orientable surface of genus
1 + 8n(1− 1/4− 1/4− 1/(4n)) = 4n− 1,
and we show that G has no faithful action on a surface of genus g smaller than this.
Any such action ofG on an orientable surface S has no reflections, since all involutions
in G are squares or products of squares and so must preserve orientation. Hence only cases
(A),(C1) or (C3) of Proposition 2.1 are applicable here.
We first consider case (A). If r = 3, then G has a generating set {u, v, w} of three
elements of orders m1,m2,m3, with uvw = 1. In the quotient G/〈x2, y2〉 ∼= D2n, two of
these three elements must have images of order 2 while the other has image of order 2n.
Without loss of generality, u and v have images of order 2, so u and v have order 4 in G,
and w must have order 2n in the quotient, so w must be of the form w = (xy)iz, where i
is relatively prime to 2n and z ∈ I . In particular, w has the same order as xy, namely 4n,
and [m1,m2,m3] = [4, 4, 4n], so the genus can only be 4n − 1. Similarly, if r = 4, then
G is generated by four elements t, u, v, w of orders m1,m2,m3,m4 such that tuvw = 1,
but again, at least two of these generators must have order at least 2 in the dihedral quotient
G/〈x2, y2〉 and order at least 4 inG, which gives g ≥ 1+8n(2−1/2−1/2−1/4−1/4) =
4n+ 1 > 4n− 1, a contradiction.
In case (C1), the group G is generated by two elements d and u such that um1 =
(d2u)m2 = 1. The images of u and d in the dihedral quotient G/〈x2, y2〉 ∼= D2n have
orders 2 and 2, or 2 and 2n, in some order, and since n ≥ 2 it follows that in G these two
elements both have order at least 4. Thusm1 ≥ 4. The same argument applies to d and ud2,
so m2 ≥ 4 as well. Thus 1/m1 + 1/m2 ≤ 1/2, and so g = 1 + 8n(1− 1/m1 − 1/m2) ≥
1 + 8n/2 > 4n− 1, a contradiction.
In case (C3), the group G is generated by two elements d1 and d2 such that d12d22 = 1.
Now d1 and d2 both have order greater than 2 (since G is not cyclic or dihedral), but in the
dihedral quotient G/〈x2, y2〉, they must both have order 2 (since d12 = d2−2), and hence
278 Ars Math. Contemp. 4 (2011) 271–289
both have order 4 in G. It follows that the central subgroup 〈d12〉 of G has order 2, with
dihedral quotient of order 8n. The latter quotient, however, is also generated by the images
of x and y, and since 4n > 4 the only way this can happen is for those images to both have
order 2. But then x2 and y2 are both equal to d12, which is impossible (since 〈x2, y2〉 has
order 4).
Thus we can do no better than genus 4n− 1, as claimed. 
For the case n = 1, we note that V1 is generated by its elements d1 = x and d2 = xy,
with d12d22 = x2x2 = 1. Since also the subgroup generated by x2 and y has index 2
but contains neither x nor xy, it follows that V1 acts with signature (2;−; [−], {−}) on an
orientable surface of genus g = 1 (that is, on the torus), and so σ(V1) = 1, rather than
4n−1 = 3. On the other hand, it is well known that the simple group L2(7) has symmetric
genus 3, and hence we have shown that for every positive integer g ≡ 3 mod 4, there exists
a group of symmetric genus g.
The same is known to be true for all g ≡ 1 mod 4, since May and Zimmerman showed
that the abelian group (C2)3 × C2m of order 16m has symmetric genus 4m+ 1, for every
positive integer m; see [27]. We can add to these results the following:
Theorem 3.2. σ(Un) = 4n− 3 for every positive integer n.
Proof. The proof of this theorem is largely similar to that of the previous one. LetG = Un,
and let I be the set of elements of order 1 or 2 in G, namely the eight elements of the form
z or (xy)nz with z ∈ {1, x2, y2, x2y2}. Note that the images of all these elements in the
dihedral quotient G/〈x2, y2〉 ∼= D2n are central.
The generating set {x, y, (xy)−1} gives an action with signature (0; +; [4, 4, 2n]; {−})
on an orientable surface of genus
1 + 8n(1− 1/4− 1/4− 1/(2n)) = 4n− 3.
When n = 1 this action is toroidal, and as U1 has no faithful action on the sphere (by
Maschke’s theorem), we may assume that n > 1. We now suppose that G has a faithful
action on some surface S of genus smaller than 4n− 3.
IfG acts on S without reflections, then the same kind of arguments as made in the proof
of Theorem 3.1 lead only to contradictions, whether S/G is orientable or non-orientable.
If, instead, G acts with reflections, then the images of all involutions in G are central in
the dihedral quotient G/〈x2, y2〉 ∼= D2n, so there must be more than one non-involution in
any generating set for G. This implies that we have case (B2) of Proposition 2.1, but then
m1 and m2 must be at least 4, and so
g ≥ 1 + 8n(1− 1/4− 1/4) = 1 + 4n > 4n− 3.
Hence we can do no better than genus 4n− 3, and the theorem holds. 
Theorems 3.1 and 3.2 together show that the symmetric genus of a finite group can be
any odd positive integer.
4 Even genera
In this section we begin by considering another family of groups, namely
M. D. E. Conder and T. W. Tucker: The symmetric genus spectrum of finite groups 279
Wn = 〈x, y | x2 = y3n = 1, (xy)3 = y3 〉, for n ∈ Z+.
Here the relation (xy)3 = y3 implies that y3 is centralized by both y and xy, and so
is central in Wn. Thus Wn is an extension of the central cyclic subgroup 〈y3〉 by the
(2, 3, 3) triangle group 〈x, y | x2 = y3 = (xy)3 = 1 〉 ∼= A4, and in particular, this shows
that |Wn| ≤ 12n. On the other hand, the abelianisation Wn/W ′n is cyclic of order 3n
(generated by the image of y). It follows that |Wn| = 12n, with y having order 3n (and
W ′n having order 4).
Theorem 4.1. σ(Wn) = 3n− 3 for every odd positive integer n.
Proof. LetG = Wn. The generating set {x, y, (xy)−1} gives an action ofG with signature
(0; +; [2, 3n, 3n]; {−}) on an orientable surface of genus
1 + 6n(1− 1/2− 1/(3n)− 1/(3n)) = 3n− 3.
When n = 1 this action is spherical, so we will suppose that n > 1, and thatG has a faithful
action on some surface S of genus smaller than 3n− 3. Again Proposition 2.1 applies, but
since the abelianisation G/G′ is cyclic of order 3n, the group G has no subgroup of index
2, and so all surface actions of G preserve orientation. Hence we need only consider case
(A).
If r = 3, we may assume that 2 ≤ m1 ≤ m2 ≤ m3. If m1 = 2 then since G/G′ is
cyclic of odd order 3n, we would have m3 ≥ m2 ≥ 3n, giving g ≥ 3n − 3. Similarly, if
m1 = 3 then since G/〈y3〉 ∼= A4 we must have {m2,m3} = {2s, 3t} for some s, t, while
also since G/G′ ∼= C3n, both s and t must be divisible by n; hence one of m2 and m3 is at
least 2n while the other is at least 3n, giving g ≥ 1 + 6n(1− 1/3− 1/(2n)− 1/(3n)) =
4n − 4 > 3n − 3. Next, m1 cannot be 4 or 5, since y3 has odd order and G/〈y3〉 ∼= A4,
which has no element of order 4 or 5. Finally, if m1 ≥ 6 then g ≥ 1 + 6n(1− 1/6− 1/6−
1/6) = 3n+ 1 > 3n− 3. Hence the case r = 3 is impossible.
With r = 4, we require two of the mi to be 2 (otherwise ξ ≥ 2 − 1/2 − 3/3 = 1/2),
but then the other two must be divisible by 3n since G has abelianisation of odd order 3n,
so g ≥ 1 + 6n(2− 2/2− 2/(3n)) = 6n− 3 > 3n− 3, which again is impossible. 
If G is a finite group and H is a subgroup of G, then clearly σ(H) ≤ σ(G). Moreover,
if σ(H) = g and G has a faithful action on a closed orientable surface of genus g, then
σ(G) = σ(H). These observations give us two further families of groups of symmetric
genus divisible by 6.
Corollary 4.2. For every odd positive integer n, each of the following groups has symmet-
ric genus 3n− 3:
(a) 〈u, v | u4 = v3n = (uv)2 = 1, u−1v3u = v−3 〉, which is an extension of Cn by
S4, of order 24n, and
(b) 〈 a, b, c | a2 = b2 = c2 = (ab)4 = (bc)3n = (ac)2 = [a, (bc)3] = 1 〉, which is an
extension of Cn by S4 × C2, of order 48n.
Proof. If G is the group in (a), then the subgroup H generated by x = u2 and y has index
2 in G, and as the relation (uv)2 = 1 implies that xy = u2v = uv−1u−1, we see that xy is
conjugate to v−1 = y−1, and further, (xy)3 = uv−3u−1 = v3 = y3. An easy application
of Reidemeister-Schreier theory (see [22] for example) now shows that H has the same
280 Ars Math. Contemp. 4 (2011) 271–289
presentation as Wn, so |G| = 24n. The quotient of G by the cyclic normal subgroup 〈v3〉
is the (2, 3, 4) triangle group S4, and the generating set {uv, u, v} gives an orientation-
preserving action of G with signature (0; +; [2, 4, 3n]; {−}) on an orientable surface of
genus 1 + 12n(1− 1/2− 1/4− 1/(3n)) = 3n− 3 as well.
Similarly, if L is the group in (b), then the subgroup generated by u = ab and v = bc
has index 2 in L, and has the same presentation as G, so |L| = 48n. The quotient of
L by the cyclic normal subgroup 〈(bc)3〉 is the extended (2, 3, 4) triangle group S4 × C2
(the full symmetry group of the regular octahedron), and the generating set {c, a, b} gives
a reflective action of G with signature (0; +; [−]; {(2, 4, 3n)}) on an orientable surface of
genus 1 + 24n(1/2− 1/4− 1/8− 1/(6n)) = 3n− 3. 
Our next two theorems use direct products for genera congruent to 4 mod 6.
Theorem 4.3. σ(Cn × S4) = 3n+ 1 for every odd integer n ≥ 5, n 6= 9.
Proof. Let G = Cn × S4, where n is odd and n ≥ 5. In this group of order 24n, let u =
(1, (1, 3)), v = (1, (1, 2), and w = (z, (1, 2)(3, 4)), where z is a generator for Cn. Then
u, v and w generate G, since w2 generates Cn while the images of wu = (z, (1, 2, 3, 4))
and v in G/Cn generate S4. Also u, v and w satisfy u2 = v2 = [u,w] = [v, w]2 = 1, and
the index 2 subgroup Cn × A4 of G contains w but neither u nor v. It follows that G has
a reflective action with signature (0; +; [−]; {(1), (2)}) on an orientable surface of genus
1 + 12n(1/2− 1/4) = 3n+ 1; this is case (B3) of Proposition 2.1.
Now suppose G has a faithful action on a surface S of genus g smaller than 3n + 1.
Then g < 1 + |G|/8, so we consider the cases of Proposition 2.1 with ξ < 1/4. Note that
the abelianisation of G is C2n, generated by the image of uw or vw, and the centre of G is
Cn, with quotient G/Z(G) ∼= S4.
Consider first case (A). If r = 3 and one branch point has order 2, then by considering
the projections onto Cn and S4 we see that the other two must have orders 3n and 4n,
giving g ≥ 1 + 12n(1 − 1/2 − 1/(3n) − 1/(4n)) = 6n − 6 > 3n + 1. Similarly, if
one branch point has order 3, then the other two must have orders at least 2n/3 and 4n/3,
giving g ≥ 1 + 12n(1− 1/3− 3/(2n)− 3/(4n)) = 8n− 26 > 3n+ 1 for n > 5, while
for the remaining case n = 5 the other two must have orders at least 2n and 4n, giving
g ≥ 1 + 12n(1− 1/3− 1/(2n)− 1/(4n)) = 8n− 8 > 3n+ 1. If all branch points have
order 4 or more, then g ≥ 1 + 12n(1 − 1/4 − 1/4 − 1/4) = 3n + 1, but abelianisation
shows this bound cannot be attained when n > 2. On the other hand, if r = 4, then at least
two branch points must have order 2, but then the orders of the other two are divisible by
n, so g ≥ 1 + 12n(2− 1/2− 1/2− 1/n− 1/n) = 12n− 23 > 3n+ 1.
In case (B1), the group G is generated by an element of order m and s involutions,
and ξ = s/2 − 1/m −
∑s
j=1 1/(2qj) < 1/4. Abelianisation gives m ≥ n ≥ 5, which
forces s = 1, so the signature is (0; +; [m]; {(q)}). It follows that G is generated by two
elements u and v such that u2 = vm = [u, v]q = 1, with v lying in Cn × A4 and u lying
outside it. Projection onto S4 shows the images of u and v must be a 2-cycle and a 3-cycle,
with also the image of [u, v] a 3-cycle. Hence m is divisible by 3 and q = 3. But now
g ≥ 1 + 12n(1/2 − 1/n − 1/6) = 4n − 11 > 3n + 1 for all n > 12, while for the
remaining cases n = 5, 7 and 11 (not divisible by 3), abelianisation givesm ≥ 3n and then
g ≥ 1 + 12n(1/2− 1/(3n)− 1/6) = 4n− 3 > 3n+ 1.
In case (B2), the group G is generated by three elements u, v and c such that um1 =
vm2 = c2 = [c, uv]q = 1. The condition ξ < 1/4 implies 1/m1 + 1/m2 > 3/4 and so
M. D. E. Conder and T. W. Tucker: The symmetric genus spectrum of finite groups 281
(without loss of generality) m1 = 2, but then G cannot have abelianisation of order 2n. In
cases (B3) and (C2), we have ξ = 1/2 − 1/(2q) < 1/4, which forces q = 1, but this is
impossible (by Proposition 2.1) since G/Z(G) is neither cyclic nor dihedral. Case (C3) is
ruled out for the same reason.
Finally, in case (C1) the group G is generated by two elements d and u such that
um1 = (d2u)m2 = 1. As in case (B2), the condition ξ < 1/4 gives 1/m1 + 1/m2 > 3/4,
so (without loss of generality) m1 = 2, and then in the abelianisation the image of d2 must
have order n, so the image of d2u has order 2n, giving m2 ≥ 2n. This, however, implies
g ≥ 1 + 12n(1− 1/2− 1/(2n)) = 6n− 5 > 3n+ 1, another contradiction. 
It is easy to prove that the symmetric group S5 has symmetric genus 4; indeed S5 and
S5 × C2 are the only finite groups of symmetric genus 4; see Section 6. To cover the
remaining cases of genus 3n+ 1 for n = 3 and 9, we have the following:
Theorem 4.4. σ(C3 × C3 × C3n) = 18n− 8 for every positive integer n.
Proof. This is similar to previous theorems, but does not use Proposition 2.1. First, let
u, v, w be respective generators of the three factors of of G = C3 × C3 × C3n, and
z = (uvw)−1. Then taking {u, v, w, z} as generating set gives an orientation-preserving
action ofGwith signature (0; +; [3, 3, 3n, 3n]; {−}) on a surface of genus 1+(27n/2)(2−
1/3−1/3−1/(3n)−1/(3n)) = 18n−8. Clearly, no orientation-preserving action can do
better, since G has rank 3 and adding extra generators simply increases the genus. The fact
that G has only one involution can be used to eliminate potentially better cases where the
action does not preserve orientation. We leave the details to the reader. See also [24, 27]. 
We note that the same argument can be used to show that
σ(Cm × Cm × Cmn) = 1 +m2((m− 1)n− 1) for all odd m and all n.
When m > 3, however, this is not very useful for filling gaps in the symmetric genus
spectrum; for example, when m = 5 it covers only genera congruent to 76 mod 100.
Our final theorem in this section begins to cover the gaps for genus g ≡ 2 mod 6. For
this, we take the following family of groups:
Tn = 〈x, y | x2 = y3n = [x, y3] = 1, (xy)3 = (yx)3 〉, for n ∈ Z+.
We note that the abelianisation of Tn is C2 × C3n, and it follows that the order of y is
exactly 3n. Also y3 is central, and T/〈y3〉 ∼= T1.
The group T1 has order 48, with derived subgroup T1′ isomorphic to the quaternion
group Q8. To see this, observe that the subgroup of index 2 in T1 generated by X = xyx
and Y = y has presentation 〈X,Y | X3 = Y 3 = 1, XY X = Y XY 〉 and is isomorphic
to SL(2, 3) via the mapping X 7→
(
0 1
−1 −1
)
and Y 7→
(
−1 1
−1 0
)
. (In fact T1
and SL(2, 3) are two of the three groups that have (White) genus 1 but not symmetric genus
1, as given in [18, Section 6.4]; see our comments just before Theorem 6.2.) Here [X,Y ]
maps to
(
−1 1
1 0
)
, which has order 4 and together with its conjugates generates a
subgroup isomorphic to Q8.
Thus |Tn| = n|T1| = 48n, and also Tn′ (of index 6n in Tn) is isomorphic to Q8, for
all n. It follows that the commutator of every two elements of Tn has order 1, 2 or 4, and
282 Ars Math. Contemp. 4 (2011) 271–289
the commutator of every generating pair for Tn has order 4. In particular, [x, y] (= X−1Y )
has order 4, and z = [x, y]2 is a central involution. Furthermore, the subgroupK generated
by y3 and z is central in Tn, with quotient PSL(2, 3) ∼= A4. Since A4 has trivial centre, it
follows that K = Z(Tn) and Tn/Z(Tn) ∼= A4.
Theorem 4.5. σ(Tn) = 9n− 7 for every odd positive integer n.
Proof. Let G = Tn, where n is odd. Since the index 2 subgroup generated by y and xyx
contains y but not x, the relations y3n = x2 = [x, y]4 = 1 imply that G has a reflective
action with signature (0; +; [3n]; {(4)}) on an orientable surface of genus
1 + 24n(1/2− 1/(3n)− 1/8) = 9n− 7.
Now suppose G has a faithful action on a surface S of genus g < 9n − 7. Again we
can use Proposition 2.1 (since |Tn| = 48n > 4(9n− 7)).
First consider case (A). If r = 3 and u, v, w are the corresponding generators for G
of orders m1 ≤ m2 ≤ m3 and satisfying uvw = 1, then it is not difficult to verify (with
the help of MAGMA [3] for example) that the images of u, v, w in the quotient T1 must
have orders 2, 3, 12, or 2, 6, 12, or 3, 12, 12, or 4, 12, 12, or 6, 12, 12. If m1 = 2, then
abelianisation shows that m2 ≥ 3n andm3 ≥ 12n. Thus g ≥ 1+24n(1−1/2−1/(3n)−
1/(12n)) = 12n−10 > 9n−7. Similarly, ifm1 = 3, 4 or 6 thenm3 ≥ m2 ≥ 12n, giving
g ≥ 1 + 24n(1 − 1/3 − 1/(12n) − 1/(12n)) = 16n − 3 > 9n − 7, and if m1 ≥ 7 then
g ≥ 1+24n(1−1/7−1/7−1/7) = (96/7)n+1 > 9n−7. Similarly, if r = 4 then at least
two branch points must have order 2, but then the orders of the other two branch points must
be divisible by 3n (by abelianisation), so g ≥ 1+24n(2−1/2−1/2−1/(3n)−1/(3n)) =
24n− 15 > 9n− 7. Hence no such action gives genus smaller than 9n− 7.
In case (B1), the group G is generated by an element of order m and s involutions, and
ξ = s/2− 1/m−
∑s
j=1 1/(2qj) < 1/2. Abelianisation gives m ≥ 3n, which then forces
s = 1 (since we may suppose qj ≥ 2 for all j when s > 1). Thus G is generated by two
elements u and v such that u2 = vm = [u, v]q = 1, and necessarily m ≥ 3n and q ≥ 4 (by
our comments above about Tn′), so we can do no better than our specified action of genus
9n− 7.
In case (B2), the group G has generators u, v and c such that um1 = vm2 = c2 =
[c, uv]q = 1, and g = 1+24n(3/2−1/m1−1/m2−1/q) ≥ 1+24n(1−1/m1−1/m2).
In order to make this less than 9n − 7 we require 1/m1 + 1/m2 > 15/24, so without
loss of generality m1 is 2 or 3. But if m1 = 2 then abelianisation gives m2 ≥ 3n, so that
g ≥ 1 + 24n(1− 1/2− 1/(3n)) = 12n− 7 > 9n− 7. Similarly, if m1 = 3 then m2 ≥ n,
and so for n > 1 we have g ≥ 1 + 24n(1− 1/3− 1/n) = 16n− 23 > 9n− 7, while for
n = 1 we have m2 ≥ 2 (since the orders of interior branch points are always at least 2) and
then g ≥ 1 + 24(1− 1/3− 1/2) = 5 > 9n− 7.
In case (B3), the group G is generated by three elements u, v and w such that u2 =
v2 = [u,w] = [v, w]q = 1, and g = 1 + 24n(1/2− 1/2q). To make this less than 9n− 7
we require q < 4, so q = 1 or 2. Now q 6= 1 since G/Z(G) ∼= A4 is neither cyclic
nor dihedral; hence q = 2. But then [v, w] is the unique involution z in G′ ∼= Q8, so the
quotient G = G/〈z〉 is generated by u and v (of order 1 or 2) and the central element w,
which is impossible since G/Z(G) ∼= A4 is not generated by involutions.
In case (C1), the group G is generated by d and u such that um1 = (d2u)m2 = 1, and
g = 1 + 24n(1 − 1/m1 − 1/m2). For this to be less than 9n − 7 we require 1/m1 +
1/m2 > 15/24, so without loss of generality m1 is 2 or 3. But if m1 = 2 then in the
M. D. E. Conder and T. W. Tucker: The symmetric genus spectrum of finite groups 283
abelianisation C6n the image of d2 must have (odd) order 3n, so m2 ≥ 3n, which gives
g ≥ 1 + 24n(1 − 1/2 − 1/(3n)) = 12n − 7 > 9n − 7. Similarly, if m1 = 3 then in
the (cyclic) abelianisation C6n the order of the image of d2 must be n or 3n, and then so
must the order of the image of d2u (whether 3 divides n or not), and thus m2 ≥ n. This
gives g ≥ 1 + 24n(1 − 1/3 − 1/n) = 16n − 23 > 9n − 7 for n > 1, while for n = 1
we have m2 ≥ 2 (since the orders of interior branch points are always at least 2) and then
g ≥ 1 + 24(1− 1/3− 1/2) = 5 > 9n− 7.
In case (C2), the group G has two generators d and c such that c2 = [d2, c]q = 1,
and just as in the case (B3), we find q = 2. But then [d2, c] is the unique involution z in
G′ ∼= Q8, and in the quotient G = G/〈z〉, the image of d2 is central, so G/Z(G) ∼= A4 is
generated by the involutory images of c and d, which is again impossible.
Finally, the case (C3) is ruled out also by the fact that G/Z(G) ∼= A4 is neither cyclic
nor dihedral (or since G has no abelian subgroup of index 2). 
The first theorems in this section cover all genera g ≡ 0 or 4 mod 6, and the last one
covers all g ≡ 2 mod 18. All that remains are genera g ≡ 8 or 14 mod 18, and thus we
have proved the first part of Theorem 1.2.
5 Filling some gaps
Let A be any finite abelian group that admits an automorphism θ of order 3, with the
property that for some x ∈ A, the element x and its images xθ and xθ2 generate A, and
satisfy xxθxθ
2
= 1. Now form the semi-direct product Aoθ C6, in which conjugation by
a generator y of C6 induces the automorphism θ on A. This group has order 6|A|, and the
element y3 is a central involution (so Aoθ C6 ∼= (Aoθ C3)×C2). Also xy−1 has order 6,
since (xy−1)3 = x(y−1xy)(y−2xy2)y−3 = xxθxθ
2
y3 = y3, which has order 2; similarly
xy−2 has order 3. Hence the elements x1 = xy−2, x2 = y and x3 = yx−1 = (xy−1)−1
satisfy the relations x13 = x26 = x36 = x1x2x3 = 1, and give rise to a faithful action of
Aoθ C2 on an orientable surface of genus
1 + 3|A|(1− 1/3− 1/6− 1/6) = |A|+ 1.
There are several ways to obtain such an abelian group A and automorphism θ. For
example, if p is any prime congruent to 1 mod 3, and e is any positive integer, then the
group of units mod pe has order φ(pe) = pe−1(p − 1) ≡ 0 mod 3, so there exists a
non-trivial cube root of 1 mod pe, with 1 + λ + λ2 ≡ 0 mod pe, and then we can take
A = Cpe = 〈x |xp
e
= 1 〉 and θ the automorphism taking x 7→ xλ. Alternatively, if m is
any positive integer, we can take A = Cm × Cm = 〈x, z |xm = zm = [x, z] = 1 〉 and
let θ be the automorphism taking x 7→ z 7→ x−1z−1 7→ x. More generally, we have the
following:
Theorem 5.1. Let A be a non-trivial abelian group of the form Cs×Ct×Ct, where s and
t are odd, gcd(s, t) = 1, and every prime divisor of s is congruent to 1 mod 3. Then A has
an automorphism θ of order 3 with the property that σ(Aoθ C6) = |A|+ 1.
Proof. For each maximal prime-power divisor qi = peii of s, there exists a non-trivial cube
root λi of 1 mod qi, with 1 + λi + λ2i ≡ 0 mod qi, and then by the Chinese remainder
theorem, there exists an integer λ such that λ ∼= λi mod qi for all i, from which it follows
284 Ars Math. Contemp. 4 (2011) 271–289
that also 1 + λ + λ2 ≡ 0 mod s. Now if {u, v, w} is the standard basis for A, we may
define the automorphism θ by setting
uθ = uλ, vθ = w, wθ = v−1w−1.
It is easy to see that θ has order 3, and also that if x = uv, we have
xxθxθ
2
= (uv)(uλw)(uλ
2
v−1w−1) = u1+λ+λ
2
vwv−1w−1 = 1.
Moreover, since gcd(s, t) = 1, the element x = uv generates Cs × Ct, and so x and
xθ = uλw generate A. Hence by the observations made earlier, the group G = A oθ C6
has a faithful action on a surface of genus |A|+ 1.
Now suppose G has a faithful action on a surface S of genus g < |A| + 1. Again
Proposition 2.1 applies.
Here the fact that G ∼= (A oθ C3) × C2 has a unique involution z (with non-cyclic
quotient G/〈z〉 ∼= Aoθ C3) implies that any generating set for G must contain at least two
elements of order greater than 2. In particular, this rules out cases (B1), (B3) and (C2).
Next consider case (A). If r = 3, let u, v, w be the corresponding generators for G of
orders m1,m2,m3 and satisfying uvw = 1. Then since any two of u, v, w generate G,
their orders must all be at least 3, and since G/A ∼= C6, at least two must be 6 or more, so
g ≥ 1 + 3|A|(1− 1/3− 1/6− 1/6) = |A|+ 1. Similarly if r = 4 then at least two of the
mi are at least 3, so g ≥ 1 + 3|A|(2− 1/2− 1/2− 1/3− 1/3) ≥ |A|+ 1.
In case (B2), where G has generators u, v and c satisfying um1 = vm2 = c2 =
[c, uv]q = 1, we have m1,m2 ≥ 3, so g ≥ 1 + 3|A|(1− 1/3− 1/3) = |A|+ 1. Similarly,
in case (C1), the group G is generated by d, u and v such that um1 = vm2 = d2uv = 1.
Here we have G = 〈d, u〉 = 〈d, d2u〉, so both u and v have order greater than 2, which
again gives g ≥ 1 + 3|A|(1− 1/3− 1/3) = |A|+ 1.
Finally, case (C3) is impossible since G has no abelian subgroup of index 2. 
Note that a similar construction was used in [12, §3.4.1] for groups acting on non-
orientable surfaces.
Corollary 5.2. Let g be an even positive integer, such that g − 1 = p1e1p2e2 . . . pmem
where the pi are distinct primes greater than 3 (and the ei are positive integers). If pei 6≡ 5
mod 6 for 1 ≤ i ≤ m, then there exists a finite group of symmetric genus g.
Proof. By the given condition, we can write g − 1 as st2, where all prime divisors of s are
congruent to 1 mod 6, and then σ(Cs × Ct × Ct) = g by the above theorem. 
Note that this completes the proof of Theorem 1.2, since if g ≡ 8 or 14 mod 18 then
every prime divisor of g − 1 must be congruent to ±1 mod 6.
Corollary 5.3. Let A be as in Theorem 5.1. Then there exists a semi-direct product A :
(C6 × C2) having symmetric genus |A|+ 1.
Proof. Instead of θ, take the automorphism ψ of A = Cs × Ct × Ct given by
uψ = uξ, vψ = w, wθ = v−1w,
where ξ is an integer satisfying ξ2 − ξ + 1 ≡ 0 mod s (so that ξ3 ≡ −1 mod s), whose
existence is guaranteed by the assumption that pei−1 ≡ 0 mod 6 for every maximal prime-
power divisor peii of s. Now form the semi-direct product G = A : (C6 × C2) by letting a
M. D. E. Conder and T. W. Tucker: The symmetric genus spectrum of finite groups 285
generator y of the factor C6 induce the automorphism ψ on A and letting a generator z of
the factor C2 centralize A. In this group G, take the elements
x1 = uvy
3z, x2 = y, and x3 = (uvy4z)−1.
Since z centralizes A and conjugation by y3 inverts every element of A, we have x12 =
uvu−1v−1 = 1, and similarly (uvy4z)3 = u1+ξ
2+ξ4(v(v−1w)w−1)y6z3 = u1−ξ+ξ
2
z =
z, so x1, x2 and x3 have orders 2, 6 and 6 respectively. Moreover, the subgroup generated
by x1, x2 and x3 contains y = x2 and z = x33 and so contains also uv = x1zy3 and uξw =
(uv)y , so contains all of G. Thus G has a faithful action on an orientable surface of genus
1+6|A|(1−1/2−1/6−1/6) = |A|+1. Finally, since 1+ξ2+(ξ2)2 ≡ 1−ξ+ξ2 ≡ 0 mod
s, the automorphism ψ2 is the same as θ or θ−1, so the index 2 subgroup of G generated
by A and y2z or y4z is isomorphic to the group A oθ C6 considered in Theorem 5.1, and
it follows that σ(G) = σ(Aoθ C6) = |A|+ 1. 
6 Remaining gaps in the spectrum
Our constructions so far cover well over eight ninths of all positive integers as possibilities
for σ(G) for some finite group G. The smallest genera not covered by Theorem 1.2 are
g = 86 = 5× 17 + 1, g = 116 = 5 × 23 + 1, and g = 188 = 11 × 17 + 1. There are,
however, no gaps in the spectrum at these three values of g, for :
(a) the group 〈x, y | x8 = y51 = 1, x−1yx = y25 〉 ∼= C51 o25 C8 of order 408 (and
with centre of order 3) has symmetric genus 86, realised by an action with signature
(0; +; [2, 24, 24]; {−});
(b) the group 〈u, v, t | u25 = v5 = [u, v] = t4 = 1, t−1ut = u19v4, t−1vt = u15v2 〉,
which is a semi-direct product (C25 × C5) : C4 of order 125, has symmetric genus
116, realised by an action with signature (0; +; [4, 4, 25]; {−}); and
(c) the group 〈x, y | x10 = y99 = 1, x−1yx = y17 〉 ∼= C99 o17 C10 of order 990 (and
with trivial centre) has symmetric genus 188, realised by an action with signature
(0; +; [2, 10, 45]; {−}).
These examples (which were found with the help of MAGMA [3]) are members of
further infinite families that might be helpful in closing the gaps, but the orders of groups
in such infinite families do not behave as well as those considered in this and the previous
two sections.
It is illuminating to consider the groups of small symmetric genus. We have used
MAGMA [3] to determine all finite groups of symmetric genus 2 to 32 inclusive, and May
and Zimmerman [30] have done this independently for genus 2 to 8. Table 1 presents a
summary of the groups of symmetric genus 2, 8, and 14, which we felt could be helpful in
finding further families of groups of symmetric genus congruent to 2 mod 6.
The family of metabelian groups is helpful in dealing with some genera. We have found
(again with the help of MAGMA [3]) that there are no metacyclic groups of symmetric
genus 2, 3, 4, 5, 6, 10, 13, 24, 26, 30 or 37. On the other hand, although there are no
metabelian groups of symmetric genus 2, 3 or 4, there do exist abelian or metabelian groups
of symmetric genus g for all g between 5 and 38. This provides evidence (however flimsy
that it might be) in support of the following stronger version of Conjecture 1.3:
Conjecture 6.1. For every integer g ≥ 5, there is a finite abelian or metabelian group G
with symmetric genus σ(G) = g.
286 Ars Math. Contemp. 4 (2011) 271–289
σ |G| Description of G Signature(s)
2 24 SL(2, 3) (0;+; [3, 3, 4]; {−})
48 GL(2, 3) (0;+; [2, 3, 8]; {−}), (0;+; [−]; {(3, 3, 4)})
48 SL(2, 3) : C2 (0;+; [3]; {(4)})
96 GL(2, 3) : C2 (0;+; [−]; {(2, 3, 8)})
8 42 (C7 o2 C3)× C2 (0;+; [3, 6, 6]; {−}), (1;−; [3, 3]; {−}), (0;+; [3, 3]; {(1)})
48 SL(2, 3) : C2 (0;+; [3, 4, 8]; {−})
84 C7 : (C6 × C2) (0;+; [2, 6, 6]; {−}), (0;+; [3]; {(2, 2)}), (0;+; [2, 3]; {(1)})
96 GL(2, 3) : C2 (0;+; [−]; {(3, 4, 8)})
336 PSL(2, 7)× C2 (0;+; [−]; {(3, 3, 4)}), (0;+; [3]; {(4)})
672 PGL(2, 7)× C2 (0;+; [−]; {(2, 3, 8)})
14 78 (C13 o3 C3)× C2 (0;+; [3, 6, 6]; {−}), (1;−; [3, 3]; {−}), (0;+; [3, 3]; {(1)})
120 SL(2, 5) (0;+; [3, 4, 5]; {−})
156 C13 : (C6 × C2) (0;+; [2, 6, 6]; {−}), (0;+; [3]; {(2, 2)}), (0;+; [2, 3]; {(1)})
156 C13 o2 C12 (1;−; [2, 3]; {−})
240 SL(2, 5) : C2 (0;+; [−]; {(3, 4, 5)})
1092 PSL(2, 13) (0;+; [2, 3, 7]; {−})
2184 PGL(2, 13) (0;+; [−]; {(2, 3, 7)})
2184 PSL(2, 13)× C2 (0;+; [−]; {(2, 3, 7)})
Table 1: Groups of symmetric genus σ(G) = 2, 8 or 14
Aside from the symmetric genus, the strong symmetric genus and the symmetric cross-
cap number, there is another parameter known simply as the genus (or sometimes the White
genus). The (White) genus γ(G) of a finite group G is the smallest non-negative integer
g such that some Cayley graph for the group G has a 2-cell embedding in an orientable
surface of genus g; see [33].
In general, γ(G) ≤ σ(G), but computation of γ is far more difficult than σ(G), except
for very smallG, those with γ(G) = 0 or 1, and thoseG attaining the lower bound σ(G) ≥
1+ |G|/168 when γ(G) > 1; see [32]. The reason for this is that Cayley graphs have many
possible embeddings into surfaces, very few of which allow the vertex-transitive action
of the group on the Cayley graph to induce an action of the same group on the surface. In
particular, it is difficult to say much at all about the spectrum of values of the (White) genus
function γ.
Here we make just one observation:
Theorem 6.2. If g = st2 + 1 for integers s and t with t ≥ 2, then there exists a group of
(White) genus g.
Proof. Let G be the abelian group C2st × C2t × C2. Take two copies of the standard
embedding of a 4-valent Cayley graph for C2st × C2t = 〈x, y | x2st = y2t = [x, y] = 1 〉
in the torus (where edges correspond to multiplication by x±1 or y±1, and every face is a
quadrilateral corresponding to a commutator [x±1, y±1]). The quadrilaterals with opposite
corners (x2i, y2j) and (x2i+1, y2j+1) form a 2-factor in the Cayley graph, and these can
be used to tube together the two copies, giving an all-quadrilateral embedding of a Cayley
graph for G. (This is a simple example of the White-Pisanski construction; see [18] or
[31].) The genus of the resulting surface is 1 + (2t)(2st)(1 − 5/2 + 5/4) = 1 + st2.
Furthermore, since the only way to have any triangular faces is to allow generators of order
3, which would increase the valence to 7 or more, this all-quadrilateral embedding has the
M. D. E. Conder and T. W. Tucker: The symmetric genus spectrum of finite groups 287
least possible genus among all embeddings of Cayley graphs for G. Thus γ(G) = 1 + st2.

Corollary 6.3. There exists a group of (White) genus g whenever g ≡ 1 mod t2 for some
non-zero integer t. In particular, this occurs whenever g ≡ 1 mod 4 or 9.
The results presented in this paper are also related to the study of regular maps on sur-
faces. Roughly speaking, a regular map is a 2-cell embedding of a graph or multigraph into
a closed surface with a large group of symmetries (automorphisms of the graph or multi-
graph that preserve the faces), so that there exists automorphisms inducing a single-step
local rotation about any given vertex or any given face. As such, regular maps generalise
the Platonic solids (which may be viewed as regular maps on the sphere) to surfaces of
higher genera.
On orientable surfaces, a regular map is called reflexible if it admits an automorphism
fixing an edge (and so reversing orientation), and chiral otherwise. Some very recent re-
search has considered the genus spectrum of these and other classes of regular maps. In
[2], for example, some ground-breaking work by Breda, Nedela, and Širáň [2] showed that
there are no regular maps at all on non-orientable surfaces of Euler characteristic−p where
p > 13 is prime and p ≡ 1 mod 12. We took a different approach in [13], to obtain a
shorter proof of the latter, and also to prove that:
(a) there is no chiral regular map on an orientable surface of genus p + 1, where p is
prime and p 6≡ 1 mod 6, 8, 10, and
(b) there is no reflexible regular map with simple underlying graph (that is, with no loops
or multiple edges) on an orientable surface of genus p+ 1, where p > 13 is a prime
congruent to 1 mod 6.
The precise genus spectra of each of these two classes of maps are still unknown.
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