I'm missing a detail in the following proof of the Second Sylow Theorem:
Let G be a finite group and p a divisor of $|G|$. Let $|G| = p^km$ with $p \nmid m$. Then if $H \le G$ is a p-subgroup then there exists $P \le G$ p-subgroup such that $H \le P$.
The proof begins with a lemma which goes as follows (I put it here for completeness):
Let $P$ be a p-subgroup of Sylow of $G$ and $H \le G$ a p-subgroup such that $H \le N_G(P)$ then $H \le P$.
The proof of theorem goes as follows:
Let $S$ be the set of p-subgroups of Sylow and let's take the conjugation action of $G$ on $S$. Let $P_1 \in S$ be a p-subgroup and $T = O(P_1)$ be the orbit of such p-subgroup then I understand the following facts:
- $P_1 \trianglelefteq N_G(P_1) \le G$ and therefore $[G:P_1] = [G:N_G(P_1)][N_G(P_1):P_1]$
but I don't understand why
- $[G:P_1] = m$
which should lead to the fact that $p \nmid |T| = [G:N_G(P_1)]$.
I'd say that it is really simple but I'm stuck, thanks for your answers.
Note that $P_1$ was chosen from the set of $p$-Sylow subgroups, so
$$[G:P_1]=\vert G\vert / \vert P_1\vert$$
And as $P_1$ is a $p$-Sylow, $\vert P_1\vert=p^k$, so that lets you what you want.
The point of $p\not\mid [G:N_G(P_1)]$ is because $P_1\leq N_G(P_1)$, so $\vert N_G(P_1)\vert= p^kr$ with $1\leq r\leq m$, so again:
$$[G:N_G(P_1)]=\vert G\vert / \vert N_G(P_1)\vert=m/r$$
That is prime to $p$.