Normalizers/centralizers of Sylow 2 subgroups of subgroups of a finite solvable group with elementary abelian sylow subgroup

Question

Let $G$ be a finite solvable group with an abelian Sylow $p$ subgroup $S$.

A classic theorem of Burnside says that $S$ has a normal complement if and only if the Centralizer of $S$ in $G$ is equal to the Normalizer of $S$ in $G$. Since the centralizer is contained in the normalizer we can just look at the index. So for any group $G$ let ${\rm icn}_p(G)$ be the index of the centralizer of a Sylow $p$ subgroup in its normalizer.

First just consider $p=2$, but much of this, if not all, could be independent of $p$. I would like to know how ${\rm icn}$ behaves w.r.t. subgroups.

If $G$ is solvable with abelian Sylow $2$ subgroup $S$ is it true for every subgroup $H \subset G$ that ${\rm icn}(H) \le {\rm icn}(G)$? Moreover is it true that ${\rm icn}(H)$ is a divisor of ${\rm icn}(G)$?

These seem like they should be elementary, but I don't see how to approach.

A quick computation shows that if solvability and abelian are not required, then the second statement about divisors is false, but the first inequality still seems to hold. An example of this is the symmetric group $S_4$. Here the Sylow 2 subgroup is not abelian, but for each subgroup of $K$ of $S_4$, $icn(H)$ is still not greater than 4. In fact $icn(S_4) = 4$, and $A_4 \subset S_4$ gives $icn(A_4) = 3$

Is this true for any finite group $G$ and subgroup $H \subset G$ that ${\rm icn}(H) \le{\rm icn}(G)$?

Thanks for your help

Answer 1

What you are looking for is the idea of control of fusion. Let $S$ be a Sylow $p$-subgroup of $G$, and let $H$ be a subgroup of $G$ containing $S$. We say that $H$ controls fusion in $G$ with respect to $S$ if, whenever $A$ and $B$ are two subsets of $S$ that are conjugate in $G$ via some element $g$, then there exists $h\in N_G(S)$ such that $A^h=B$ and, furthermore, conjugation by $g$ and $h$ induce the same function on $A$.

Burnside proved that if $S$ is abelian then $N_G(S)$ controls fusion in $S$ with respect to $G$. In general, if $N_G(S)$ controls fusion then $N_H(T)/C_H(T)$ is isomorphic to a subgroup of $N_G(S)/C_G(S)$, whenever $H$ is a subgroup and $T$ is a Sylow $p$-subgroup of $H$. So your second claim holds whenever $N_G(S)$ controls fusion. This is really a statement about $p$-groups: a $p$-group $S$ where $N_G(S)$ always controls fusion when $S$ is a Sylow in $G$ are called resistant. Almost all finite $p$-groups are resistant.

Counterexamples to your general claim, that $\mathrm{icn}(H)\leq \mathrm{icn}(G)$, abound for simple groups, particularly for the prime $2$, when $\mathrm{icn}(G)$ can equal $1$. In this case, Frobenius's normal $p$-complement theorem states that $G$ has a normal $p$-complement if and only if $\mathrm{icn}(H)$ is a $p$-group for all subgroups $H$. If we choose a group without a normal $2$-complement, but with $N_G(S)=SC_G(S)$, then we should be able to find a counterexample.

For a concrete one, let $G=\mathrm{SL}_2(9)$ and $H=\mathrm{SL}_2(3)$ for $p=2$. Then $\mathrm{icn}(G)$ has order $8$, because $N_G(S)=S$ and the group is quaternion. The Sylow $2$-subgroup of $H$ is also quaternion but of order $8$. However, it is normal in $H$, so $\mathrm{icn}(H)=|H|/2=12$.