In a forthcoming paper with Venkatesh and Westerland, we require the following funny definition. Let G be a finite group and c a conjugacy class in G. We say the pair (G,c) is nonsplitting if, for every subgroup H of G, the intersection of c with H is either a conjugacy class of H or is empty.
For example, G can be the dihedral group of order 2p and c the class of an involution.
The case where c is an involution is o special interest to us. One way to construct nonsplitting pairs is by taking G to be a semidirect product of N by (Z/2^k Z), where N has odd order, and c is the conjugacy class containing the involutions of G. Are these the only examples? In other words:
Question 1: Is there a nonsplitting pair (G,c) with c an involution but where the 2-Sylow subgroup of G is not cyclic?
Slightly less well-posed questions:
Question 2: Are there "interesting" examples of nonsplitting pairs with c not an involution? (The only example we have in mind is G = A_4, with c one of the classes of 3-cycles.)
Question 3: Does this notion have any connection with anything of pre-existing interest to people who study finite groups?
Update: Very good answers below already — I should add that, for maximal "interestingness," the conjugacy class c should generate G. (This eliminates the examples where c is central in G, except in the case G = Z/2Z).
Best Answer
Answer: I cheated and asked Richard Lyons this question (or at least, the reformulation of the problem, conjecturing that (G,c) is nonsplitting for an involution c with
<c>generating G if and only if there exists an odd A such that G/A = Z/2). His response:Good question! This is a famous (in my circles) theorem - the Glauberman Z^
*-Theorem. (Z^*(G) is the preimage of the exponent 2 subgroup of the center of G/O(G), and O(G)=largest normal subgroup of G of odd order.)Z^
*-Theorem: If c is an involution of G then c\in Z^*(G) iff [c,g] has odd order for all g\in G iff for any Sylow 2-subgroup S of G containing c, c is the unique G-conjugate of itself in S.The last property is absolutely fundamental for CFSG. The proof uses modular character theory for p=2. Attempts to do it with simpler tools have failed.
George Glauberman, Central Elements in Core-free Groups, Journal of Algebra 4, 1966, 403-420.
Older Remarks:
Comment 1: Suppose that P = Z/2+Z/2 is a 2-Sylow. If x lies in P, then P clearly centralizes x, and thus the order of
<x>divides #G/P, and is thus odd. By a theorem of Frobenius, G has an odd number of elements of order 2, and thus we see it has an odd number of conjugacy classes of elements of order 2. Yet, by the Sylow theorems, every element of order 2 is conjugate to an element of P. If c lies in P, then by nonsplitting, it is unique in its G-conjugacy class in P. Thus there must be exactly three conjugacy classes of elements of order 2, and thus no element of P is G-conjugate. By a correct application of Frobenius' normal complement theorem, we deduce that G admits a normal subgroup A such that G/A \sim P. Yet<c>generates G, and thus the image of<c>generates G/A. Yet G/A is abelian and non-cyclic, a contradiction.Comment 2: Suppose that A is a group of order coprime to p such that p | #Aut(A). Let G be the semidirect product which sits inside the sequence:
1 ---> A ---> G --(phi)--> Z/pZ --> 0;
Let c be (any) element of order p which maps to 1 in Z/pZ. If c is conjugate to c^j, then phi(c) = phi(c^j). Hence c is not conjugate to any power of itself.
Let H be a subgroup of G containing c (or a conjugate of c, the same argument applies). The element c generates a p-sylow P of H (and of G). It suffices to show that if gcg^-1 lies in H, then it is conjugate to c inside H. Note that gPg^-1 is a p-Sylow of H. Since all p-Sylows of H are conjugate, there exists an h such that gPg^-1 = hPh^-1, and thus h c^j h^-1 = gcg^-1. Yet we have seen that c^j is not conjugate to c inside G unless j = 1. Thus gcg^-1 = hch^-1 is conjugate to c inside H.
I just noticed that you wanted
<c>to generate G. It's not immediately clear (to me) what condition on A one needs to impose to ensure this. Something like the automorphism has to be "sufficiently mixing". At the very worst, I guess, the group G' generated by<c>still has the property, by the same argument.This works more generally if p || G and no element of order p is conjugate to a power of itself. (I think you know this already if p = 2.)
The case where the p-Sylow is not cyclic is probably trickier.
Examples: A = (Z/2Z)+(Z/2Z), p = 3. (This is A
_4).A = Quaternion Group, p = 3. (This is GL
_2(F_3) = ~A_4, ~ = central extension).A = M^37, M = monster group, p = 37.