Until someone suggests better terminology, let me call a subgroup H of a finite group G segregated if every class function on H can be extended to a class function on G. Equivalently, H should have the property that any two of its elements which happen to be conjugate in G should be conjugate in H.

(This seems to have something to do with the known notions of fusion of conjugacy classes for subgroups, but whenever I've looked in the literature on such things I've not quite found what I was after.)

Here are the few things I know right now:

  1. An obvious example where this happens is the usual inclusion of $S_m \hookrightarrow S_{m+n}$ where we think of $S_m$ as permutations of $\{1,\dots, m+n\}$ that fix each element of $\{m+1,\dots, m+n\}$.

  2. An obvious example where this doesn't happen is the inclusion $A_m \subset S_m$ since one can always find even permutations of the same cycle type which are not conjugate in $A_m$.

  3. If H is contained in Z(G) then it is a segregated subgroup of G.

  4. Malnormal subgroups (i.e. Frobenius complements) are segregated; thus one can have abelian segregated subgroups which aren't central.

Question 1. Suppose G is a finite group with a proper subgroup that is non-abelian. Does it contain a proper subgroup that is non-abelian and segregated?


Question 2. Suppose G is a finite group with non-trivial centre. Does it contain a proper subgroup which is non-abelian and segregated?

These questions are motivated by the study of certain Banach algebra norms one can put on the algebra of class functions (with pointwise product), and certain invariants one can associate to these Banach algebras, which do not increase if one passes to quotient algebras. So one would like to take ${\mathcal C}\ell$(G) and restrict to a subgroup H, but this only works well for what I want if the image of the restriction map is all of ${\mathcal C}\ell$(H), i.e. when H is segregated in the sense described above.


2 Answers 2


I think that the group ${\rm SL}(2,3)$ shows that the answer to both questions is"no" in general. It has a proper non-Abelian subgroup (quaternion of order $8$) and a non trivial center. Its only proper non-Abelian subgroup is quaternion of order $8,$ but that is not segregated, since all its elements of order $4$ are conjugate in ${\rm SL}(2,3)$ but not within the quaternion subgroup. It does look as if the answer will "usually" be yes though, but I am not sure how to make this precise.

LATER EDIT: Here is a (well-known, though maybe not in this terminology and context) general "fusion and transfer" type result which seems relevant. Let $P$ be Sylow $p$-subgroup of a finite group $G$. Then $P$ is segregated if and only if $G$ has a normal $p$-complement (ie normal subgroup $K$ of order prime to $p$ with $G = KP).$ One way round is clear (normal $p$-complement implies segregation). For the other direction one could use Frobenius's normal $p$-complement theorem and induction, but it's perhaps quicker to use a Theorem of Tate (and D.G. Higman's Focal Subgroup Theorem). For (using the focal subgroup theorem), the fact that $P$ is segregated implies that $P \cap [G,G] = [P,P].$ Then Tate's theorem implies that $G$ has a normal $p$-complement.

EVEN LATER EDIT: In fact, using transfer, it is possible to prove that if $G$ is a finite non-trivial perfect group (ie $G = [G,G] \neq 1$) and $H$ is a Hall subgroup of $G$ (ie a subgroup whose order and index are coprime), but $H$ is not perfect, then $H$ is not segregated. So, for example, $A_{4}$ should not be segregated in $A_{5},$ and indeed it is not, whereas $S_{3}$ and $D_{10}$ are segregated in $A_{5}.$

  • $\begingroup$ Thanks for the quick answer! Is there a "geometric" reason why the order 4 elements are conjugate in SL(2,3)? $\endgroup$
    – Yemon Choi
    Apr 20, 2014 at 22:20
  • $\begingroup$ I'm not sure what you would consider a "geometric" reason. There are any number of algebraic expanations, including the fact that $Q_{8}$ admits ${\rm Sp}(2,2) \cong {\rm SL}(2,2)$ as a group of automorphisms. $\endgroup$ Apr 20, 2014 at 22:35
  • 2
    $\begingroup$ I think there are similar examples (for both Quesions 1 and 2) that are extensions of extraspecial $p$-groups of exponent $p$, for odd primes $p$, by odd order cyclic groups of odd order dividing $p+1$. There is one of order $375 = 5^3\times 3$ for instance. I could not think of any other types of examples. $\endgroup$
    – Derek Holt
    Apr 21, 2014 at 8:23

This does not really address the two questions, but here is an example where segregated subgroups arise fairly naturally. Let $A \subseteq {\rm Aut}(G)$, where $|A|$ and $|G|$ are coprime. Let $C = {\bf C}_G(A)$, the fixed point subgroup. Then $C$ is necessarily segregated in $G$. In fact, the map $K \mapsto K \cap C$ defines a bijection from the set of $A$-fixed classes of $G$ onto the set of all classes of $C$.

  • $\begingroup$ Thanks! As is mentioned rather vaguely at the end of my question, locating non-abelian segregated subgroups would be useful wherever possible, so I will look to see whether the examples given by your observation occur "in the wild" when studying my problem. $\endgroup$
    – Yemon Choi
    Apr 24, 2014 at 21:41
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    $\begingroup$ It does occur in the wild in groups of Lie type. For example, ${\rm GL}(2,3^{5})$ admits a (coprime) field automorphism of order $5$ with fixed point subgroup ${\rm GL}(2,3)$ which is segregated, via Marty's observation. In fact, this does point out another general source of segregated subgroups. If $F \subseteq K$ are finite fields, then ${\rm GL}(n,F)$ is segregated in ${\rm GL}(n,K)$ ( via the theory of the rational canonical form). It is the case "in nature" that (eg) finite simple groups have relatively few coprime automorphisms, though this isn't a priori obvious, needing CFSG. $\endgroup$ Apr 25, 2014 at 8:54

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