5
$\begingroup$

I'm stuck on this problem that I've found in Isaacs' Finite Group Theory. I've tried thinking about it for a while but didn't came up with a solution, so I'm asking for some hints or even for a full solution if you wish, everything is welcomed since I don't see any way to follow.

The problem is the following.

Let $G$ be a finite group and $P\in \text{Syl}_p(H)$, where $H\subseteq G$, meaning that $P$ is a Sylow $p$ group in $H$. Let $N_G(P)$ be the normalizer of $P$ in $G$, and let $N_G(P)\subseteq H$. Then show that $p\nmid|G:H|.$

$\endgroup$
10
$\begingroup$

Let $P$ be a $p$-subgroup of $H$ whose normalizer is completely contained in $H$. Write $|P|=p^n$. If $p$ divides $[G:H]$, then $G$ contains a $p$-subgroup of order $p^{n+1}$ that contains $P$. Call this $Q$. Since $[Q:P]=p$, $P$ is maximal in $Q$ hence normal in $Q$, so $Q\subseteq N_G(P)\subseteq H$. But this means that $H$ contains a $p$-subgroup of order strictly larger than that of $P$, so $P$ is not a $p$-Sylow subgroup of $H$.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

We actually have $|G:H| \equiv 1 \pmod{p}$ in this case. Indeed, considering the action of $P$ on the set of left cosets $G/P$ we get $$ |G:P| \equiv |\mathbf{N}_G(P) : P|\pmod{p}$$ since $P$ is a $p$-group. Now since $\mathbf{N}_G(P) \subseteq H$, $|\mathbf{N}_G(P) : P|$ is not divisible by $p$ hence we can cancel this term in the above equivalence and get $$ |G : \mathbf{N}_G(P)| \equiv 1 \pmod{p}$$ (This comes for free if you show $P \in \text{Syl}_p(G)$ first, since $|G:\mathbf{N}_G(P)| = n_p(G)$)

Furthermore $|H:\mathbf{N}_G(P)| =n_p(H) \equiv 1 \pmod{p}$ and hence $$|G:H| \equiv 1 \pmod{p}$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.