The Strong Symmetric Genus and Generalized Symmetric Groups: Results and a Conjecture

The strong symmetric genus and generalized symmetric groups: results and a conjecture

Michael A. Jackson Grove City College, USA [email protected] Special Session on Group Theory, Actions, and Computation AMS-NZMS Joint Meeting Wellington, New Zealand December, 15 2007

1 Strong Symmetric Genus

Given a ﬁnite group G, the smallest genus of any closed orientable topological surface on which G acts faithfully as a group of orientation preserving symmetries is called the strong symmetric genus of G.

The strong symmetric genus of the group G is denoted σ0(G).

By a result of Hurwitz [1893], if σ0(G) > 1 for a ﬁnite group 0 |G| G, then σ (G) ≥ 1 + 84 .

2 Known results on the strong symmetric genus

All groups G such that σ0(G) ≤ 4 are known. [Broughton, 1991; May and Zimmerman, 2000 and 2005]

For each positive integer n, there is exists a ﬁnite group G with σ0(G) = n. [May and Zimmerman, 2003]

3 Known results on the strong symmetric genus continued

The strong symmetric genus is known for the following groups:

• PSL2(q) [Glover and Sjerve, 1985 and 1987]

• SL2(q) [Voon, 1993]

• the sporadic ﬁnite simple groups [Conder, Wilson and Woldar, 1992; Wilson, 1993, 1997 and 2001]

4 Known results on the strong symmetric genus continued

The strong symmetric genus is known for the following groups:

• alternating and symmetric groups [Conder, 1980 and 1981]

• the hyperoctahedral groups [J, 2004]

• the remaining ﬁnite Coxeter groups [J, 2007]

5 The Generalized Symmetric Groups

G(n, m) = Zm o Σn for n > 1 and m ≥ 1.

This group can also be described as the group of n × n matrices which contains both the permutation matrices and the diagonal matrices with entries in a multiplicative cyclic group of size m.

G(n, 1) is the traditional symmetric group Σn. G(n, 2) is the hyperoctahedral group Bn.

Now we will consider the generalized symmetric groups of type G(n, m), where m ≥ 3 .

6 Triangle Groups and Actions

Deﬁnition 1 Let p, q, and r be positive integers. We de- ﬁne the (p, q, r) triangle group by the following presentation T (p, q, r) = ha, b, c|ap = bq = cr = abc = 1i.

Given a triangle group T (p, q, r) we get a triangle with the π π π angles p, q , and r .

1 1 1 Such a triangle exhibits elliptic geometry if p + q + r > 1, 1 1 1 Euclidean geometry if p + q + r = 1, or hyperbolic geometry 1 1 1 if p + q + r < 1.

7 Triangle Groups and Actions continued

Tessellate the appropriate plane with the given triangle and pick a particular triangle.

By labeling the vertices of the triangle as a, b, and c cor- responding to the angles, we can say that the action a, b, 2π 2π 2π and c are rotation by angles of p , q , and r about the cor- responding vertices in the correct directions.

This gives an action of T (p, q, r) on the appropriate plane. (We will call this P).

8 Triangle Groups and Actions continued

Now the fundamental domain of this action is just two copies of our triangle. So we see that P/T (p, q, r) is topologically a sphere.

Now assume that G is a ﬁnite group which is a quotient of T (p, q, r). Let K be the kernel of the map T (p, q, r) → G and assume that K is torsion free.

9 Triangle Groups and Actions continued

Then G acts on some triangulated surface S = P/K such that S/G = P/T (p, q, r) is also a sphere. So we get the map π : S → S/G with three branch points of orders p, q, and r.

P −→ S = P/∆ ↓ π↓ =∼ S/G = P/T (p, q, r) −→ S/G

10 Riemann-Hurwitz Equation

We want to ﬁnd the genus of S. This genus can be found since S is a branched cover of the sphere S/G. We will use the Riemann-Hurwitz formula which says that |G| 1 1 1 genus(S) = 1 + (1 − − − ). 2 p q r

So given a finite group G to find the strong symmetric genus σ0(G) one may try to find a triangle group having G as a quotient by finding a pair of generators for the group G.

11 Generating Pairs

If a ﬁnite group G has generators x and y of orders p and q respectively with xy having the order r, then we say that (x, y) is a (p, q, r) generating pair of G.

By the obvious symmetries concerning generators, we will use the convention that p ≤ q ≤ r.

12 Maximal Generating Pairs

Following the convention of Marston Conder, we say that a (p, q, r) generating pair of G is a minimal generating pair if there does not exist a (k, l, m) generating pair for G with 1 1 1 1 1 1 k + l + m > p + q + r .

If G has a (p, q, r) generating pair, it is the quotient of the triangle group T (p, q, r).

Recall that the groups of small strong symmetric genus are well known. So generally we will assume that σ0(G) ≥ 2 or equivalently that any generating pair will be a (p, q, r) gener- 1 1 1 ating pair with p + q + r < 1.

13 Riemann-Hurwitz Equation Applied

Using the Riemann-Hurwitz equation, we see that given any generating pair of G, we get an upper bound on the strong symmetric genus of G. If G has a (p, q, r) generating pair, then |G| 1 1 1 σ0(G) ≤ 1 + (1 − − − ). 2 p q r

14 Requirements for orders of generating pairs.

Lemma 2 Let G = G(n, m) be a generalized symmetric group with n > 2 and suppose that x, y form a (p, q, r)-generating pair of G.

– For any m at least two of p, q, and r are even.

– If m is even all three numbers, p, q, and r are even.

– If k > 2 is a prime power (st = k with s prime) such that k divides m, then k divides at least two of p, q, and r.

15 A Lemma by Singerman

Lemma 3 (Singerman) Let G be a ﬁnite group such that σ0(G) > 1. If |G| > 12(σ0(G) − 1), then G has a (p, q, r) generating pair with 1 1 1 1 σ0(G) = 1 + |G| · (1 − − − ). (1) 2 p q r

Notice that Singerman’s lemma implies that if G has a mini- 1 1 1 5 mal (p, q, r) generating pair such that p + q + r ≥ 6, then the strong symmetric genus is given by this generating pair.

This works well for the standard symmetric groups and the hyperoctahedral groups. However, for generalized symmetric groups G(n, m) where m > 2 we need a diﬀerent version of this lemma.

16 An Extension of Singerman’s Lemma

Theorem 4 (J, An extension of Singerman’s Lemma) Let G = G(n, m) be a generalized symmetric group with n > 3. Suppose k > 2 is a prime power (st = k with s prime) such 2k 0 that k divides m. If |G| > k−2(σ (G)−1), then G has a (p, q, r) generating pair with 1 1 1 1 σ0(G) = 1 + |G| · (1 − − − ). 2 p q r

Combining this theorem with the previous discussion reduces the problem of ﬁnding the genus of a generalized symmetric group G(n, m) to ﬁnding the minimal generating pair of type 1 1 1 2 (p, q, r) for the group G(n, m) provided that p + q + r > k for some prime power k > 2 dividing m.

17 More on the last theorem

Let G = G(n, m) be a generalized symmetric group with n > 3. If G(n, m) has a minimal (2, q, r) generating pair and m is divisible by a prime power k ≥ 4, then the strong symmetric genus is given by the this minimal generating pair. This is 1 2 true since 2 ≥ k.

This only leaves the cases m = 3 and m = 6.

For m = 3 it will be enough to see that G(n, 3) has a minimal 1 1 2 (2, 3, r) generating pair since 2 + 3 > 3.

For m = 6 it will be enough to see that G(n, 6) has a minimal 1 1 2 (2, 6, r) generating pair since 2 + 6 = 3.

18 A look at triples The triples in the order previously discussed which satisfy the requirements for a hyperbolic generating pair of G(n, m): (2, 3, 8) (2, 4, 5) (2, 3, 10) (2, 3, 12) = (2, 4, 6) (2, 3, 14) (2, 3, 16) (2, 4, 7) (2, 3, 18) (2, 3, 20) (2, 3, 22) (2, 3, 24) = (2, 4, 8) (2, 3, 26) (2, 3, 28) (2, 3, 30) = (2, 5, 6) (2, 3, 32) (2, 3, 34) (2, 3, 36) = (2, 4, 9) (2, 3, 38) (2, 3, 40) (2, 3, 42) (2, 3, 44) (2, 3, 46) (2, 3, 48) (2, 3, 50) (2, 3, 52) (2, 3, 54) (2, 3, 56) (2, 3, 58) (2, 3, 60) = (2, 4, 10) (2, 3, 62) ... (2, 3, 132) (2, 4, 11) (2, 3, 134) ... (2, 4, 12) = (2, 6, 6) = (3, 4, 4) (2, 5, 8) (2, 4, 14) . . . Marston Conder has shown that for n > 167 each symmetric group Σn = G(n, 1) has a minimal (2, 3, 8)-generating pair.

19 A look at triples continued The triples in the order previously discussed which satisfy the requirements for a hyperbolic generating pair of G(n, m) for an even m: (2, 4, 6)

(2, 4, 8)

(2, 4, 10)

(2, 4, 12) = (2, 6, 6) (2, 4, 14) . . . The author has shown that for n > 8 each hyperoctahedral group Bn = G(n, 2) has a minimal (2, 4, 6)-generating pair.

20 A look at triples continued The triples in the order previously discussed which satisfy the requirements for a hyperbolic generating pair of G(n, m) with m divisible by 3: (2, 3, 12) (2, 3, 18) (2, 3, 24) (2, 3, 30) (2, 3, 36) (2, 3, 42) (2, 3, 48) (2, 3, 54) (2, 3, 60) ... (2, 3, 132) ... (2, 6, 6) . . . The author has shown that for n > 33 each generalized symmetric group of type G(n, 3) has a minimal (2, 3, 12)- generating pair.

21 A look at triples continued The triples in the order previously discussed which satisfy the requirements for a hyperbolic generating pair of G(n, m) with m divisible by 4:

(2, 4, 8)

(2, 4, 12) . . . But no general symmetric group of type G(n, 4) has a minimal (2,4,8)-generating pair. Why is this?

22 An Additional Requirement

There is one more requirement for generating pairs of generalized symmetric groups.

Proposition 5 If the generalized symmetric group G(n, m) has a (2, m, r)-generating pair, then

– if m is divisible by 4, then r ≥ 3m and m divides r.

– if m is divisible by 2 but not by 4, then r ≥ 2m and m divides r.

– if m is odd, then r ≥ 4m and 2m divides r.

This proposition follows from the wreath product structure of the generalized symmetric groups.

23 Explanation of last proposition

Consider the short exact sequence and elements:

n (Zm) ,→ G(n, m) Σn

a 7−→ [σ, a] 7−→ σ

b 7−→ [τ, b] 7−→ τ Suppose that [σ, a] has order m and [τ, b] has order r, such that their product [στ, τ −1(a)b] has order 2.

Then τ −1(a)b must be order 2 or 1. Thus if m is odd, or is divisible by 4, both a and b must have order divisible by m.

Since [σ, a] has order m, a must be ﬁxed by σ which means that there must be a cycle of τ which non-trivially permutes b and has a length at least 3.

24 Explanation of last proposition continued

Since there must be a cycle of tau which non-trivially permutes σ and has a length of at least 3. r must be at least 3m.

If m is odd, r must be even. Therefore, r is at least 4m. If m is divisible by 4, r can be 3m or larger.

In the case of m ≡ 2(mod 4, the 2 factor of m may be put in τ −1(a)b and a 3 Thus the order of [τ, b] need only be at least 2m but since it must also be even it must be at least 2m.

25 A look at triples continued The triples in the order previously discussed which satisfy the requirements for a hyperbolic generating pair of G(n, m) with m divisible by 4:

(2, 4, 8)

(2, 4, 12) . . . It appears that for n > 15 each generalized symmetric group of type G(n, 4) has a minimal (2, 4, 12)-generating pair.

26 Review of Results

[Conder] For n > 167 each symmetric group Σn = G(n, 1) has a minimal (2, 3, 8)-generating pair and has a strong symmetric genus of

0 n! σ (Σn) = + 1. 48 [J] For n > 8 each hyperoctahedral group Bn = G(n, 2) has a minimal (2, 4, 6)-generating pair and has a strong symmetric genus of n 0 n! 2 σ (Bn) = + 1. 24 [J] For n > 33 each generalized symmetric group of type G(n, 3) has a minimal (2, 3, 12)-generating pair and has a strong symmetric genus of n! 3n σ0(G(n, 3)) = + 1. 24

27 Additional Information and Conjectures

It appears that for n > 15 each generalized symmetric group of type G(n, 4) has a minimal (2, 4, 12)-generating pair and has a strong symmetric genus of n! 4n σ0(G(n, 4)) = + 1. 12 Conjecture: For n > 20 each generalized symmetric group of type G(n, 5) has a minimal (2, 5, 20)-generating pair

Conjecture: For n > 18 each generalized symmetric group of type G(n, 6) has a minimal (2, 6, 12)-generating pair.

Conjecture: For n > 21 each generalized symmetric group of type G(n, 7) has a minimal (2, 7, 28)-generating pair.

28 Conjecture Part I

Conjecture 6 Let m ≥ 3 be an integer. Consider the generalized symmetric groups of type G(n, m). There exists an integer N, such that for any n > N the group G(n, m) has a minimal (2, m, r)-generating pair where

– if m ≡ 0(mod 4), then r = 3m

– if m ≡ 2(mod 4), then r = 2m

– if m is odd, then r = 4m.

29 Conjecture Part II

Recalling the discussion around the lemma of Singerman and its generalization we get the following conjecture, which would be a corollary to the previous conjecture.

Conjecture 7 Let m ≥ 3 be an integer. Consider the generalized symmetric groups of type G(n, m). There exists an integer N, such that for any n > N the group G(n, m) has following strong symmetric genus:

0 n! mn 2m−5 – If m is odd, σ (G(n, m)) = 2 4m + 1.

0 n! mn 3m−8 – If m ≡ 0(mod 4), σ (G(n, m)) = 2 6m + 1.

0 n! mn m−3 – If m ≡ 2(mod 4), σ (G(n, m)) = 2 2m + 1.