5

This is exercise $1.D.5$ of Isaacs' "Finite Group Theory". It goes:

Let $G$ be a finite group and let $H$ be a subgroup of $G$. Suppose $N_G(P) \subset H$ for all $p$-subgroups $1 \neq P$ of $H$. Then, $H$ is a Frobenius complement in $G$.

For the relevant definition, a Frobenius complement is a proper subgroup $H$ such that $H \cap H^g = 1$ for all $g \in G \setminus H$.

The problem above comes with a hint, namely:

HINT: Show that $H \cap H^g$ satisfies the hypothesis for $g \in G$. Then, suppose it is not-trivial, take a non-trivial Sylow subgroup $Q \in \operatorname{Syl}_p(H \cap H^g)$ and prove $Q, Q^{g^{-1}}$ are conjugate in $H$

I have been able to prove the first part of the hint, but I can't get the second part. Here's what I have thus far:

Let $Q$ be as in the hint. In particular, $Q\subset H$ and $Q^{g^{-1}} \subset H \cap H^{g^{-1}}$.

By a former problem, we can extend $Q \subset P$, where $P \in \operatorname{Syl}_p(H)$ is such that $P \cap (H \cap H^g) = Q$. The same way, we can define a $P'$ with $P' \cap (H^{g^{-1}} \cap H) = Q^{g^{-1}}$.

By the Second Sylow Theorem, there exists $h \in H$ such that $P^h = P'$. Therefore, $P^h \cap (H \cap H^{gh}) = Q^h \implies Q^h = P'\cap (H \cap H^{gh})$

I don't know how to continue. I hoped the $h$ given above (by the SST) would be the one to work, but the two expressions didn't come out equal...

How should I proceed?

Thanks in advance!

Gauss
  • 2,969

1 Answers1

5

Thats what I did. In your notation let $Q\in Syl_p(H\cap H^g)$ and $P\in Syl_p(H)$ such that $Q\le P$. Since $Z(P)\le N_G(Q)$, it follows that $Z(P)\le H\cap H^g$. But $P\le N_G(Z(P))$, so $P\le H\cap H^g$ (notice, it means that $Q=P$). Consequently, $P^{g^{-1}}\le H$. And we are done, since for some $h\in H$ we have $P^{g^{-1}h}=P$, so $g^{-1}h\in N_G(P)$ and $g^{-1}\in N_G(P)\le H$, a contradiction.

pansh1n
  • 434