Generalized Symmetric Groups with the Best Strong Symmetric Genus

Generalized symmetric groups with the best strong symmetric genus

Michael A. Jackson Grove City College [email protected]

Special Session on Computational Group Theory AMS Central Section Meeting Kalamazoo, Michigan October 18, 2008

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.

The strong symmetric genus has also been found for the groups G(n, 3) [J, 2008] and for the groups G(n, m) with n = 3, 4 and 5. [Ginter, Johnson, McNamara, 2008]

6 Generating Pairs and Riemann-Hurwitz

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.

The existance of a (p, q, r) generating pair gives a faithful orientation preserving action of the group G on a surface S. The genus of the surface S is found from the Riemann- Hurwitz formula: |G| 1 1 1 genus(S) = 1 + (1 − − − ). 2 p q r By the obvious symmetries concerning generators, we will use the convention that p ≤ q ≤ r.

7 Minimal Generating Pairs

We will use the convention that a (p, q, r) generating pair of G is called a minimal generating pair if no generating pair for the group G gives an action on a surface of smaller genus.

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. (Hyperbolic generating pair)

With these assumptions a (p, q, r) generating pair is minimal 1 1 1 if p + q + r is the largest of any generating pair.

8 Riemann-Hurwitz Equation and Requirements of generating pair

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

Lemma 1 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. Then • 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. 9 A Lemma by Singerman

Lemma 2 (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 − − − ). 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. 10 An Extension of Singerman’s Lemma

Theorem 3 (J, An extension of Singerman’s Lemma) Let G = G(n, m) be a generalized symmetric group with n > 3.

• If m = 3 or m = 6 and |G| > 6(σ0(G) − 1), or

• if m > 3, m 6= 6 and |G| > 4(σ0(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

11 More on the last theorem

Combining this theorem with the previous discussion reduces the problem of ﬁnding the strong symmetric genus of a generalized symmetric group G(n, m), where m > 3 and m 6= 6, to ﬁnding the minimal generating pair of type (p, q, r) for the 1 1 1 1 group G(n, m) provided that p + q + r > 2.

If G(n, m) has a minimal (2, q, r) generating pair where m > 3 and m 6= 6, then the strong symmetric genus of G(n, m) is given by the this minimal generating pair.

12 More on the last theorem

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

Combining the theorem with the previous discussion reduces the problem of ﬁnding the genus of a generalized symmetric group G(n, m), where m = 3 or m = 6, to ﬁnding the minimal generating pair of type (p, q, r) for the group G(n, m) provided 1 1 1 2 that p + q + r ≥ 3.

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.

13 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.

14 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.

15 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. 16 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) . . . The group G(18, 4) has a minimal (2, 4, 8) generating pair. I expect all but ﬁnitely many groups of type G(n, 4) to have minimal (2, 4, 8) generating pairs. 17 General Triples

In general, the best possible generating pair for a group G(n, m) with n > 3 and m ≥ 5 allowed by basic group theory constraints is: • (2, m, m) for m even, and • (2, m, 2m) for m odd.

Goal: To show that for each m ≥ 5 there is an n such that G(n, m) has a best possible generating pair as listed above.

We will do this in three cases: m odd, m ≡ 2 (mod 4), and m ≡ 0 (mod 4).

18 Preliminaries

We create generators of G(n, m) with orders 2 and m and their product having order lcm(2, m), by ﬁnding generators of Σn with orders 2 and m and their product having order lcm(2, m) and adding the cyclic group portion to these generators.

We will call the generators of Σn, σ and τ and require the following properties: • o(σ) = m, o(τ) = 2 • τ ∈ Σn \ An • σ does not permute two points, and • σ · τ does not permute two points.

19 Preliminaries continued

If σ and τ generate Σn with the properties above, then G(n, m) is generated by elements [σ, b] and [τ, 0] where b is n n element of (Zm) and 0 is the identity of (Zm) , such that o([σ, b]) = o(σ) = m.

20 Case 1: m odd - Example G(16, 7)

This coset diagram represents the generators of Σ16 of orders 2 and 7 with a product of order 14.

21 Case 1: m odd - G(2m + 2, m)

This coset diagram is generalized for other odd m by adding points between the green lines to make each polygon an m- gon. This will make generators of Σ2m+2 of orders 2 and m with a product of order 2m. 22 Case 2: m ≡ 2 (mod 4) - Example G(22, 10)

This coset diagram represents the generators of Σ22 of orders 2 and 10 with a product of order 10.

23 Case 2: m ≡ 2 (mod 4) - G(2m + 2, m)

This coset diagram is generalized for other m ≡ 2 (mod 4) by adding an equal number of points between each pair of green lines to make each circle have m points. This will make generators of Σ2m+2 of orders 2 and m with a product of order m. 24 Case 3: m ≡ 0 (mod 4) - Example G(18, 8)

This coset diagram represents the generators of Σ18 of orders 2 and 8 with a product of order 8.

25 Case 3: m ≡ 0 (mod 4) - G(2m + 2, m)

This coset diagram is generalized for other m ≡ 0 (mod 4) by adding an equal number of points at each spot labeled A to make the polygon an m-gon and adding ”2-gons” along the loop at point B to make the total number of points 2m + 2. This will make generators of Σ2m+2 of orders 2 and m with a product of order m. 26 Conclusions

The generators of G(n, m) have been veriﬁed using MAGMA for 5 ≤ m ≤ 10, 000.

A proof that these pattern continue would be desirable.

Notice for 5 ≤ m ≤ 10, 000, we have shown that G(2m+2, m) has a minimal (2, m, lcm(2, m)) generating pair.

Looking at the results earlier in the talk. I would conjec- ture that for each m ≥ 5 there exists an Nm such that if n ≥ Nm, G(n, m) has a minimal (2, m, lcm(2, m)) generating pair.