Subgroups from which all class functions extend to class functions on the ambient group 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:


*

*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 $\{m+1,\dots, m+n\}$.

*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 $A_m$.

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

*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?

and

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.
 A: 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 $A_{5}.$ 
A: 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 $C$.
