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Let $G$ be a finite group. We let $T_{G}$ denote the set of conjugacy classes of subgroups of $G$ and let $T$ denote a set of representatives of $T_{G}$.

How to establish a bijection between the following sets:

$\{H \in T: O^{p}(H)=H\}$ and $\{S \in T: [N_{G}(S): S] \textrm{is not divisible by p} \}$.

Here $N_{G}(S)$ means the normalizer of $S$ in $G$ and $O^{p}(H)$ the smallest normal subgroup of $H$ such that $H/O^{p}(H)$ is a $p$-group.

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What's $O^p(O^p(S))$ ? – user641 Aug 31 '11 at 23:35
Notice that for the set on the right hand side, $N_G(O^p(S))$ contains $N_G(S)$; what does $S$ map to in $N_G(O^p(S))/O^p(S)$? – user641 Aug 31 '11 at 23:38
@Steve D: I don't understand your last line. Can you please explain? – user6495 Sep 1 '11 at 0:16
About what $S$ maps to? Have you thought about it? What do we know about $S/O^p(S)$? – user641 Sep 1 '11 at 0:34
If you'd prefer, I could also post a full answer - but where's the fun in that? – user641 Sep 1 '11 at 0:36
up vote 1 down vote accepted

Let $A=\{H\in T\ |\ O^p(H)=H\}$ and $B=\{S\in T\ |\ p\nmid [N_G(S):S]\ \}$. We need to define two maps $\phi:A\rightarrow B$ and $\psi:B\rightarrow A$, which are both well-defined up to conjugacy, and are inverses of one another.

The map from $B$ to $A$ is easier to describe: $\psi(S)=O^p(S)$. Note that $O^p(O^p(S))=O^p(S)$, so $\psi(S)\in A$. Since $O^p(S)^g=O^p(S^g)$, this map is well-defined.

To define $\phi(H)$, choose a subgroup $K$, with $H\le K\le N_G(H)$, such that $K/H$ is a Sylow p-subgroup of $N_G(H)/H$. Let $\phi(H)=K$. Since $N_G(H)^g=N_G(H^g)$, and Sylows are conjugate, this map is well-defined. Moreover, since $K/H$ is a p-group, $O^p(K)\subset H$, and thus $O^p(K)=H$ since $H\in A$. Thus $N_G(K)\subset N_G(H)$, and since $p\nmid [N_G(H):K]$, certainly $p\nmid [N_G(K):K]$. Thus $\phi(H)\in B$.

From what was said above, we see that $O^p(\phi(H))=H$, so that $\psi\phi(H)=H$.

In the other direction, since $O^p(S)$ is characteristic in $S$, we have $N_G(S)\subset N_G(O^p(S))$. Now $S/O^p(S)$ is a p-subgroup of $N_G(O^p(S))/O^p(S)$, and since "normalizers grow" in p-groups, yet $S\in B$, $S/O^p(S)$ must be a Sylow p-subgroup of $N_G(O^p(S))/O^p(S)$. Thus $\phi(\psi(S))=\phi(O^p(S))=S$.

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