## Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

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).

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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.

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Hott! Good thing you asked, this answer means that we weren't going to prove this on our own. – Noah Snyder Oct 29 2009 at 17:50
I suppose that's exactly the kind of thing this site was designed to do. I'd cite this as the best example so far that MO works! – Sonia Balagopalan Oct 29 2009 at 17:51
If I've followed everything correctly here's what we've shown: If c consists of involutions and (G,c) is nonsplitting and G is generated by c then G is a semidirect product of an odd order normal subgroup N by Z/2. Furthermore any such semidirect product is automatically non-splitting, the only remaining question is whether there are natural conditions on N and the involution acting on N which explain whether Z/2 generates G. – Noah Snyder Oct 29 2009 at 18:31
Cheers to FC for the important insight that a good approach to finite group theory problems is to ask a finite group theorist.... – JSE Oct 29 2009 at 20:48

Another example, different in flavour from the others: Let G be the set of affine linear transformations x --> ax+b over a finite field, and c the conjugacy class of [ax] for a \neq 1.

Proof that this works:

The conjugacy class of ax is the maps x --> ax+i, for i in Z/p.

We need to show that, if H contains ax+i and ax+j then ax+i and ax+j are conjugate in H. Since H contains ax+i and ax+j, it contains their ratio, x+(j-i)/a. Therefore, H contains every map of the form x+k. Conjugating ax+i by x+(j-i)/(a-1) gives ax+j.

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Note that this example is included in FC's example (A = additive group of the field, Z/pZ = multiplicative group of the field; OK, this isn't prime order, but I think FC's argument still applies.) Also, when c is an involution here, a = -1, and one is back in the dihedral situation. – JSE Oct 27 2009 at 4:30
I was going to say that, but then I decided it actually was slightly different. I was using Sylow to force conjugation, somehow here what is being used here is the fact that the additive group has prime order... BTW, is your involution secretly complex conjugation? – Lavender Honey Oct 27 2009 at 4:39
ALL INVOLUTIONS ARE SECRETLY COMPLEX CONJUGATION! (Well, sort of. In fact for us this involution ends up being identified with the involution on a hyperelliptic curve over a finite field; and indeed these particular double covers X -> P^1/F_q are ramified at a chosen point at infinity, so are of the type hyperellipticologists call "quadratic imaginary" -- so in that very loose sense, yeah, it's complex conjugation) – JSE Oct 27 2009 at 13:33
The classification of involutions on complex reductive Lie groups supports your claim very well, JSE. +1. – Allen Knutson May 2 2011 at 14:41

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.

FURTHER EDIT BECAUSE PREVIOUS EDIT WAS NONSENSE AND CLAIMED TO PROVE THAT P ADMITTED A NORMAL COMPLEMENT A.

If we can prove that a P-sylow admits a normal complement A, then (assuming that G is generated by <c>) we can assume that G/A, which is isomorphic to P, is generated by the image of <c>. Yet the only p-group generated by a single conjugacy class is cyclic. (Proof: P/Phi(P) is cyclic, where Phi(P) = Frattini subgroup).

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You ask for c to map to 1 in Z/p, but you never use this condition. Am I missing something? PS Nice answer! – David Speyer Oct 27 2009 at 11:38
1 is just a stand-in for "nonzero" here. – Noah Snyder Oct 27 2009 at 13:02
Suppose (G,c) is non-splitting, that c has order p and the p||#G, then you must be in FCs setting. Taking H to be the centralizer of the subgroup generated by c, nonsplitting tells you that the centralizer is equal to the normalizer. By the Burnside normal complement theorem this tells you that the subgroup generated by c has a normal complement. – Noah Snyder Oct 27 2009 at 15:53
Just because c generates G doesn't necessarily mean that c \cap P generates P, I don't think (this is what I was confused about in earlier versions of my answer below). – Noah Snyder Oct 27 2009 at 21:02
Also I think you've misunderstood the statement of Frobenius's normal complement theorem. You need that any two elements which are conjugate in P are conjugate in G, not just that this works for one conjugacy class. See: groupprops.subwiki.org/wiki/… – Noah Snyder Oct 27 2009 at 21:05
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Here's an elementary observation, in the definition of non-splitting you can restrict your attention to those H's generated by a two (not necessarily distinct) elements of c.

In particular, if no powers of elements of c are in c (for example, if c consists of involutions) and any two distinct elements of c generate all of G then it follows that (G,c) is nonsplitting.

As FC points out two involutions always generate a dihedral group. As JSE points out (G,c) for c a class of involutions is nonsplitting iff the resulting dihedral groups are all of 2*odd order. In particular, (G,c) is nonsplitting for c an involution iff the product of any two elements of c has odd order.

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Unfortunately, the only groups generated by 2 elements of order 2 are Dihedral groups. – Lavender Honey Oct 27 2009 at 21:42
Ah, good point. So "non-splitting" for involutions is just a condition on a bunch of dihedral subgroups. – Noah Snyder Oct 28 2009 at 0:41
I wonder if a Coxeter group perspective would be useful. Since G is generated by a bunch of transpositions it's a quotient of some (non-finite) Coxeter group. Is this Coxeter group also nonsplitting? – Noah Snyder Oct 28 2009 at 1:02
Re Noah's first comment here: oh, so when c is a class of involutions is non-splitting equivalent to "the product of any two members of c has odd order?" This suggests thinking about the subgroup of G generated by all products g_1g_2, with g_1, g_2 in c. I guess this is always either G or a normal subgroup of index 2? In the latter case one is surely done. – JSE Oct 28 2009 at 1:43
Sure, but what do you do with the former case? – Noah Snyder Oct 28 2009 at 4:06

Of course, if G is abelian, then the conjugacy classes of G are just the elements, and any pair (G, c) is nonsplitting. More generally, if x is in the center of G and c is the class of x, then (G, c) is nonsplitting. So the answer to Question 1 is yes. But I imagine that you're looking for more interesting examples to questions 1 and 2!

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Question 1 Here's a very simple example. Let $Q= < i,j,k >$ be the quaternion group. $-1$ is the unique involution, so is in its own conjugacy class. $(Q,-1)$ is nonsplitting, and $Q$ is its own Sylow 2-subgroup. In general, take any finite group with a unique involution. (These turn out to be cyclic, quaternion and 2 other kinds.)

I don't know what happens when you take a group with more than one involution.

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Suppose that elements of c have prime order p (for example, c consists of involutions). Let H be the subgroup of the Sylow p-group P generated by the intersection of c with P. Note that P normalizes H since H is generated by a conjugacy class in P.

Claim: H is a central cyclic subgroup of P

Proof: Let F be the Frattini subgroup of H (generated by commutators and pth powers). Since H/F is elementary abelian and generated by elements of a single conjugacy class in H (by nonsplitting), it follows that H/F is cyclic. But then by the Burnside basis theorem (a version of Nakayama's lemma for p-groups) H must also be cyclic. Since H is abelian, by nonsplitting it must be central in its normalizer (which includes P).

In fact by non-splitting H has to be central inside the normalizer of P. So we're in the situation where no two elements of c sit inside the same Sylow P. In particular the size of c divides the number of Sylow p-groups is the same as .

Studying groups where a centralizer of an involution contains the normalizer of a Sylow 2-group seems like the sort of thing the classification people might know something about. They were all about classifying groups where the centralizer of an involution has some property.

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 If c is central but P is not normal, then (G,c) is splitting but there are fewer conjugates of c then P-Sylows. In other words, although H is central in N = normalizer of P, the centralizer of H could be larger. – Lavender Honey Oct 27 2009 at 22:31 Good point. Fixed! – Noah Snyder Oct 27 2009 at 22:40 (You should also say the the size of divides the number of Sylow subgroups.) Off to an official dinner tonight, hope to finish the case P = Z/2+Z/2 during the speeches. – Lavender Honey Oct 27 2009 at 22:53

Noah wrote:

Here's an elementary observation, in the definition of non-splitting you can restrict your attention to those H's generated by a two (not necessarily distinct) elements of c.

As FC points out two involutions always generate a dihedral group. As JSE points out (G,c) for c a class of involutions is nonsplitting iff the resulting dihedral groups are all of 2*odd order. In particular, (G,c) is nonplitting for c an involution iff the product of any two elements of c has odd order.

JSE suggests looking at the subgroup generated by all pairwise products of elements in c. This either generates the whole group or generates an index 2 subgroup (which is necessarily normal and has a complement generated by any element of c). FC noted that instead of looking at pairwise products you could instead look at pairwise commutants (since the commutant of two involutions is just the square of their product and in an odd order cyclic group the square of a generator generates).

Let's concentrate on the latter case.

when does a group A admit an involution i:A-->A such that i(a)a^-1 always has odd order, and {i(a)a^-1} generates for all A generate A.

[I think the only such groups have odd order. Also, Since A has an odd number of 2-Sylow subgroups, i preserves at least one Sylow P. Yet then i(a)a^-1 lies in P, and is thus trivial. Thus i fixes P. It follows that i preserves the normalizer N of P.--FC

I'd just run through the same argument myself before realizing this is just the fact that the centralizer of an element of c in the big group contains the 2-Sylow and its normalizer (as in the Frattini answer).--Noah]

Does this have anything to do with H^1(Z/2, A)? To flesh this out, the maps Z/2->A sending the nontrivial element to i(a)a^-1 are exactly the coboundaries. --Noah

Wait a sec, since every coboundary is a cocycle (or by a direct one-line computation) if y = i(a) a^-1 then i(y) = y^-1. So in particular we'd need that A is generated by elements such that i(y) = y^-1. --Noah

Another characterization that (G,c) splits for c an involution (and <c> generates G) is that:

(i) G is generated by <c>, (ii) [g,c] has odd order for every g in G.

(the latter just says that the product (gcg^-1*c) of any two conjugates of c has odd order.) These conditions are preserved under taking quotients. Thus they hold for at least one simple group. Using the classification (urgh) I think from this one can deduce that the only simple quotient of G is Z/2Z. This would reduce the problem to the "first case" considered above. --FC.

Actually running through all involutions in all the simple groups sounds very hard to me. Is there some reason to expect that to be tractable? In particular, for the groups of Lie type? --Noah

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I might have preferred to make this a comment, but it was a little long (should I have edited my answer instead?)

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.

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 This is great, FC! – JSE Oct 28 2009 at 12:43

If your $c$ is a conjugacy class of elements of odd prime order $p$, then (using the classification of finite simple groups), it is still the case that $G = O_{p^{\prime}}(G)C_{G}(x)$ for each $x \in c$, where $O_{p^{\prime}}(G)$ denotes the largest normal subgroup of $G$ of order prime to $p$. This might be considered as an odd analogue of Glauberman's $Z^{\ast}$-theorem. If anyone could come up with a classification-free proof of such a result, it would be of considerable interest (it is relatvely easy to prove this directly when $G$ is solvable (or, more generally, $p$-solvable)). Incidentally, your question seems to be related to the old concept of pronormality (which is treated, for example, in Gorentstein's book "Finite Groups"). Back to the case $p = 2$, interesting generalizations of Glauberman's $Z^{\ast}$-theorem were given by D. Goldschmidt and also by E. Shult. One result that Shult proved which you might find interesting is in a Bull AMS paper (circa 1966), in which he showed that an element of order $p$ in a finite group which commutes with none of its other conjugates AND centralizes every $p^{\prime}$-group it normalizes, is central in $G$.

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Regarding Question 1, let G = S4, H = A4, and c = [(12)(34)]. This class does not split, and c ≅ C2×C2, which is not cyclic. I'm not sure if this is an example along the lines of your "semidirect product of N by Z/2kZ" since I forget which factor you expect to be the normal subgroup. In the example above, c is the normal subgroup, and C3 acts by inner automorphisms of c to produce A4.

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You don't get to choose H. The condition is that c must not split for any H. In your case, c splits when H is taken to be C2 x C2. – David Speyer Oct 26 2009 at 19:08
Whoops. I should read more carefully. – Sammy Black Oct 26 2009 at 22:20

Groups in which every subgroup is normal may be relevant. See http://en.wikipedia.org/wiki/Hamiltonian_group for info on such groups.

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 Is this right? In the quaternions, the conjugacy class {i,-i} splits into two conjugacy classes in the Z/4Z it generates. – JSE Oct 26 2009 at 20:35

So suppose that A is an odd order group and i is an involution of A such that elements of the form i(x) x^-1 generate A. Since A is odd order it's solvable. So consider the derived series, A, A^(1)=[A,A], A^(2)=[A^(1),A^(1)], etc. Since each of the A^(i) are characteristic subgroups the involution i restricts to each of them. Unfortunately the "non-boringness" condition doesn't seem to nicely restrict to the A^(i).

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 Some trivial observations: 1). i is linear on A/[A,A] and i(x)x^-1 generates A/[A,A], so i = -1 on A/[A,A]. 2). If A is a p-group, and <c> generates A/[A,A], then <c>` generates A. Thus A can be any p-group admitting an automorphism i such that i = -1 on A/[A,A]. – Lavender Honey Oct 30 2009 at 0:28