THE STRONG SYMMETRIC GENUS and GENERALIZED SYMMETRIC GROUPS of TYPE G(N, 3)

THE STRONG SYMMETRIC GENUS AND GENERALIZED SYMMETRIC GROUPS OF TYPE G(n, 3)

MICHAEL A. JACKSON

Abstract. The generalized symmetric groups are deﬁned to be G(n, m) = Zm o

Σn where n, m ∈ Z+. The strong symmetric genus of a ﬁnite group G is the smallest

genus of a closed orientable topological surface on which G acts faithfully as a group

of orientation preserving automorphisms. The present work extends work on the

strong symmetric genus by Marston Conder, who studied the symmetric groups,

which are G(n, 1), and by the author, who studied the hyperoctahedral groups,

which are G(n, 2). We give results for the strong symmetric genus of the groups of

type G(n, 3).

Grove City College, 100 Campus Drive, Grove City, PA 16127 [email protected]. 1 1. Introduction

The strong symmetric genus of the ﬁnite group G is the smallest genus of a compact orientable surface on which G acts as a group of orientation preserving symmetries and is denoted σ0(G). The strong symmetric genus has a long history beginning with Burnside [1]. For more general information see [10, Chapter 6]. The strong symmetric genus of many groups are known: the alternating and symmetric groups [2, 3, 4], the hyperoctahedral groups [12], the remaining ﬁnite Coxeter groups

[11], the groups PSL2(q) [8, 9] and SL2(q) [15], as well as the sporadic ﬁnite simple groups [5, 16, 17, 18].

The generalized symmetric groups G(n, m) are deﬁned for n > 1 and m ≥ 1 to be

the wreath product Zm o Σn. We notice that the traditional symmetric groups are the

generalized symmetric groups of type G(n, 1), and the hyperoctahedral groups are

the generalized symmetric groups of type G(n, 2). Since the strong symmetric genus

has been found for these two families of generalized symmetric groups, in this paper

we will ﬁnd the strong symmetric genus for the next family of generalized symmetric

groups: type G(n, 3).

We will prove the following two theorems and a corollary:

Theorem 1. For all n > 33, the generalized symmetric group G(n, 3) is a quotient

of the triangle group T (2, 3, 12) = hx, y, z|x2 = y3 = z12 = xyz = 1i.

2 Corollary 2. For all n > 33, the generalized symmetric group G(n, 3) has strong

n!3n n!2n−1 symmetric genus 24 + 1 = 8 + 1.

Theorem 3. For 2 ≤ n ≤ 33, the strong symmetric genus of the generalized

symmetric group G(n, 3) is given in Table I along with the appropriate triangle groups.

The result for G(3, 3) was found by Dustin Pierce, a student at King College.

I would like to thank the editor and referee for proving many helpful comments

and suggestions which have allowed this paper to take its present form.

2. Subgroups and generators of the generalized symmetric groups

For the present time, we will ﬁx n ≥ 3 as well as ﬁx a prime p ≥ 3, and we will look at the generalized symmetric group G(n, p). As previously stated, G(n, p)

is the wreath product Zp o Σn. G(n, p) is also as the group of all n × n generalized

permutation matrices with nonzero entries in a multiplicative cyclic group Cp. For an

element of G(n, p), we will write an ordered pair [σ, b] where σ is an element of Σn and

n b is an n-tuple of elements in Zp (i.e. an element of (Zp) ). The multiplication then

becomes [σ, b]·[τ, c] = [σ·τ, τ −1(b)+c] where addition in the n-tuple is the operation in

the group Zp performed in each position. Recall that G(n, p) is a semidirect product

n of (Zp) by Σn.

n Deﬁnition 4. For an element b ∈ (Zp) , we say that b is divisible by p if the

sum of the terms of b add to the identity of Zp. If we ﬁx a generator of Zp, we can 3 n deﬁne a group homomorphism Σ : (Zp) → Zp where the image of b is the sum of the elements in the n-tuple b. Notice that Σ is invariant under permutations of b, and that b is divisible by p if and only if Σ(b) = 0.

We now present a proposition that is well known (see [6, 13]).

Proposition 5. For n ≥ 5, let G be a subgroup of G(n, p) with π(G) = Σn; then

n−1 n G is a split extension of Σn by one of the following: 1, Zp, (Zp) , or (Zp) . In the

ﬁrst two cases, G is isomorphic to Σn and Zp × Σn respectively. In the third case,

G = {[σ, b] ∈ G(n, p)|b is divisible by p}, while G = G(n, p) in the last case.

We ﬁnish this section with a proposition that constructs generators of G(n, p) from generators of Σn

Proposition 6. Suppose σ and τ generate Σn such that o(σ) = 2 and o(τ) = p for some odd prime p. Futhermore, assume that τ ﬁxes an element i ∈ {1, 2, . . . , n}; and in the case where n ≡ 1(mod p), σ · τ has a cycle of length not divisible by p and not containing i. Let l be the length of the cycle in σ · τ that contains i. Let b = (0,..., 0) and c = (0,..., 0, 1, 0,..., 0), where the 1 is in the ith position, be

n elements in an additive group (Zp) . Under these conditions, [σ, b] and [τ, c] generate

G(n, p). In addition the elements [σ, b], [τ, c] and [σ · τ, τ −1(b) + c] have orders 2, p, and gcd(o(σ · τ), p · l) respectively. 4 Proof. Let G = h[σ, b], [τ, c]i ⊂ G(n, p). Notice that any section s :Σn → G(n, p) takes α ∈ Sn to [α, a] where a is divisible by p. Clearly [τ, c] can not be in the image of any section homomorphism and so G is not isomorphic to Σn. Similarly since [τ, c]

n−1 is not divisible by p, G is not a split extension of Σn by (Zp) by Proposition 5.

Suppose that G is a split extension of Σn by Zp. By Proposition 5, for some section homorphism s :Σn → G(n, p), G = Z(G(n, p)) × s(Σn). Letting [1, z] be a generator of Z(G(n, p)), there is a [1, d] ∈ G(n, p) such that for each [α, a] ∈ G,

−1 −1 n τ (d)+d = a+k z in (Zp) for some scalar 0 ≤ k < p. If τ ﬁxes some element j that

−1 −1 n is diﬀerent from i, then τ (d)+d 6= c+k z for all d ∈ (Zp) ; therefore n ≡ 1(mod p).

By hypothesis we can assume that σ · τ has a cycle χ of length not divisible by p and not containing i. Notice that any subset of (σ · τ)−1(d) + d−1 is divisible by p if it

−1 is closed under permutation by σ · τ. Since [σ · τ, τ (b) + c] ∈ Z(G(n, p)) × s(Σn),

τ −1(b) + c + k z = c + k z = (σ · τ)−1(d) + d−1 for some 0 ≤ k < p. Using the portion of (σ · τ)−1(d) + d−1 permuted by χ, we see that k = 0, because this portion of k · z must be divisible by p. Now considering the entirety of (σ · τ)−1(d) + d−1, we see that c + k z = c must be divisible by p, which is not possible.

Applying Proposition 5, the argument above implies that G = G(n, p). The results

−1 about the orders of [σ, b], [τ, c] and [σ·τ, τ (b)+c] follow from an easy calculation.

5 3. A generalization of Singerman’s Lemma

We begin section 3 with a result about the order of the generators of generalized

symmetric groups.

Lemma 7. Let G = G(n, p) be a generalized symmetric group with n > 3 and

p > 2 a prime. If x1, x2, x3, . . . , xn are generators of G, then at least two of the orders

of x1, x2, x3, . . . , xn and their product x1x2x3 ··· xn are divisible by p.

Proof. Recall that G(n, p) is deﬁned as a split extension as follows:

n i π 1 → (Zp) → G(n, p) → Σn → 1.

In addition we have s :Σn → G(n, p) with π ◦ s = idΣn . For each xi, we let

−1 −1 −1 −1 zi = s◦π(xi)·xi . Since π(zi) = π◦s◦π(xi)·π(xi ) = π(xi)·π(xi ) = π(xi ·xi ) = 1,

it is clear that zi is in the kernel of π. Thus zi is in the image of i. Since i is

injective, let yi be the unique preimage of zi under i. Recall the group homomorphism

n Σ:(Zp) → Zp. Since the xi’s generate G(n, p), at least one Σ(yi) must be a generator

of Zp. If Σ(yi) is a generator, then the order of xi is divisible by p. Also if only one

of the Σ(yi)’s is a generator of Zp, then Σ(y1y2 ··· yn) will also be a generator of Zp.

Thus in this case, the order of x1x2x3 ··· xn is divisible by p.

Theorem 8 (A generalization of Singerman’s Lemma (See [14])). Let G = G(n, 3) be a generalized symmetric group with n > 3. If |G| > 6(σ0(G) − 1), then G has a 6 (p, q, r) generating pair with

1 1 1 1 σ0(G) = 1 + |G| · (1 − − − ). 2 p q r

0 Proof. Let g = σ (G) and Sg be a surface of genus g with an orientation-preserving

action of G as a group of automorphisms. Because the groups of small strong symmet-

ric genus are known we know that g ≥ 2. Let Sh = Sg/G, and let (h; m1, m2, . . . , mb)

be the branching data for the covering. The Riemann-Hurwitz equation gives

" b # |G| X 1 g = 1 + 2(h − 1) + (1 − ) . 2 m j=1 j

We must show that h = 0 and b = 3. Since |G| > 6(g − 1), applying the Riemann-

Hurwitz equation immediately implies that h ≤ 1. If h = 1, again by Riemann-

Hurwitz equation, b = 0; however, this cannot happen since G is not abelian. So we know that h = 0.

From the Riemann-Hurwitz equation, it is immediate that b is either 3 or 4; however b cannot be 4 because each mj must be greater than 1 and at least two of the mj must be divisible by 3.

Corollary 9. Let G = G(n, 3) be a generalized symmetric group with n > 3. If

1 1 1 G has an (p, q, r) generating pair with (1 − p − q − r ) ≤ 1/3, then

1 1 1 1 σ0(G) = 1 + |G| · (1 − − − ) 2 p0 q0 r0

for a minimal (p0, q0, r0)-generating pair. 7 Proof. If G has an (p, q, r) generating pair, then

1 1 1 1 σ0(G) ≤ 1 + |G| · (1 − − − ). 2 p q r

1 1 1 0 If (1 − p − q − r ) ≤ 1/3, then |G| > 6(σ (G) − 1). The result follows from Theorem

4. Proofs of Theorems 1 and 3 and Corollary 2

Consider the triples (p, q, r), so that a generalized symmetric groups of type

G(n, 3) could have a minimal (p, q, r)-generating pair. Since Σn is a quotient of

G(n, 3), we know that at least two of p, q, and r must be even. In addition by

Lemma 7, we know that at least two of p, q, and r are divisible by 3. Notice that the

triple (2, 3, 12) is the best possible triple satisfying these two conditions.

Proof of Corollary 2. By Theorem 1 we have a (2, 3, 12) generating pair for G(n, 3)

when n > 33; therefore, by Corollary 9 the strong symmetric genus of G(n, 3) is given

by the (2, 3, 12) generating group when n > 33.

Proof of Theorem 1. For each n > 33, we exhibit a (2, 3, 12) generating pair for

the group G(n, 3). We start by exhibiting a (2, 3, 12) generating pair (σ, τ) of Σn.

This generating pair is exhibited using coset diagrams (see [2, 11, 12]) such that τ

ﬁxes some i ∈ {1, 2, 3, . . . , n} and that the length of the cycle of σ · τ containing i is

4. If in addition 3 6 |(n − 1), in the coset diagram we show that σ · τ has a cycle of 8 length not divisible by 3 and not containing i. Then applying Proposition 6 gives the necessary generating pair for the group G(n, 3).

The coset diagrams can be exibited individually for each 33 < n ≤ 63; beyond that, the diagrams are constructed recursively by adding an additional piece to a diagram for n − 12 (see the process in [12]). The fact that the elements (σ and τ)

in the coset diagram generate Σn is shown by demonstrating that the group they

generate contains a p-cycle z for some prime p < n − 2 such that z contains elements

a, b ∈ Γ as well as σ(a) and τ(b), where a and b may be equal (see [12]).

Proof of Theorem 3. A (2, 3, 12) generating pair can demonstrated for G(n, 3) for

n = 19, 20, 23, 26, 28, 29, 30, 31, and 32 using coset diagrams. Again this is a minimal

generating pair because it is the best possible triple satisfying the above conditions.

For the remaining groups G(n, 3), the minimal generating pair is found in Table

i. These results have been achieved using GAP [7] and by eliminating possibilities

using simple group-theoretic arguments to prove that certain coset digrams cannot

exist.

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11 n best triangle group σ0(G(n, 3)) n = 2 (2, 3, 6) 1 2·3!33 n = 3 (2, 6, 9) 18 + 1 11·4!34 n = 4 (2, 9, 12) 72 + 1 3·5!35 n = 5 (2, 6, 12) 24 + 1 4·6!36 n = 6 (2, 6, 15) 30 + 1 4·7!37 n = 7 (2, 3, 30) 60 + 1 3·8!38 n = 8 (2, 6, 12) 24 + 1 5·9!39 n = 9 (2, 6, 18) 36 + 1 4·10!310 n = 10 (2, 3, 30) 60 + 1 3·11!311 n = 11 (2, 3, 24) 48 + 1 5·12!312 n = 12 (2, 3, 36) 72 + 1 10·13!313 n = 13 (2, 3, 66) 132 + 1 4·14!314 n = 14 (2, 3, 30) 60 + 1 2·15!315 n = 15 (2, 3, 18) 36 + 1 3·16!316 n = 16 (2, 3, 24) 48 + 1 6·17!317 n = 17 (2, 3, 42) 84 + 1 2·18!318 n = 18 (2, 3, 18) 36 + 1 5·21!321 n = 21 (2, 3, 36) 72 + 1 2·22!322 n = 22 (2, 3, 18) 36 + 1 2·24!324 n = 24 (2, 3, 18) 36 + 1 3·25!325 n = 25 (2, 3, 24) 48 + 1 2·27!327 n = 27 (2, 3, 18) 36 + 1 2·33!333 n = 33 (2, 3, 18) 36 + 1 Table i. Exceptional case results for G = G(n, 3)