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: • P SL2(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 ma- trices 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 gen- erating 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 gen- eralized 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).