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

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

2009-10-26T17:36:17Z

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).

Answer by Gabe Cunningham for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-26T18:01:46Z

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!

Answer by Sammy Black for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-26T18:22:00Z

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/_2^kZ" 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.

Answer by Sonia Balagopalan for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-26T18:25:19Z

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.

Answer by Graham Jones for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-26T20:02:41Z

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

Answer by David Speyer for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-27T01:13:13Z

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.

Answer by Lavender Honey for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-27T03:38:29Z

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).

Answer by Noah Snyder for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-27T16:00:45Z

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.

Answer by Noah Snyder for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-27T16:37:06Z

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.

Answer by Lavender Honey for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-28T08:37:04Z

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.

Answer by Noah Snyder for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-28T15:39:22Z

This is a new community wiki answer which people can edit instead of writing comments.

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.

FC asks:

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

Answer by Lavender Honey for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-29T17:43:07Z

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.

Answer by Noah Snyder for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2009-10-29T20:35:02Z

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).

Answer by Geoff Robinson for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups

2011-05-02T09:12:35Z

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 $G$.