[Math] Normalizers of Sylow p-subgroups

Question

My assignment is to prove the following proposition, and I'm unsure if my proof is correct:
Let $P$ be a Sylow $p$-subgroup of $G$, and let $H$ be the normalizer of $P$ in $G$. Prove that the normalizer of $H$ in $G$ is $H$ itself (i.e. normalizers of Sylow $p$-subgroups are self-normalizing).

I argued this: $|P|=p^{\alpha}$, and $P$ is normal in $H$, since $P$ is a subgroup of the normalizer of $P$ in $G$. Since $P$ is a subgroup of $H$, $|H:P|= 1$ (case 1) or $m$ (case 2).

$H$ is a subgroup of $G$, so $|H|$ divides $|G|$, therefore we have $|H:P||G:H|= 1$ (following from case 1 from above), or $m$ (following from case 2 from above).

Now I analyzed each case:

1)
$1*|G:H|=m \implies |H|=|G|/m \implies |H|=p^{\alpha} \implies H=P$, and we already know that the normalizer of $P$ in $G$ is $H$, so if $H=P$, the normalizer of $H$ in $G$ is $H$.

2)
$m*|G:H|=m \implies |G:H|=1 \implies G=H$, so obviously the normalizer of $H$ in $G$ is the same as the normalizer of $H$ in $H$, which is obviously all of $H$.

Is this legitimate?

Answer 1

In general: let $S \in Syl_p(G)$ and $H \leq G$, with $N_G(S) \subseteq H$, then $N_G(H)=H$. Moreover, $[G:H]\equiv 1$ mod $p$.

Proof for the first part it suffices to show that $N_G(H) \subseteq H$: observe that $S \in Syl_p(H)$ and take a $g \in N_G(H)$. Then $S^g \subseteq H$ and hence $S^g=S^h$ for some $h \in H$. That means $gh^{-1}\in N_G(S)$, so $g \in hN_G(S) \subseteq H$.

For the second part we use the fact that in general for a $p$-subgroup $P$ of $G$, it holds that $[G:P]\equiv[N_G(P):P]$ mod $p$ (this can be shown by letting $G$ act by right multiplication on the right cosets of $P$). Further, since $N_G(S) \subseteq H$, $N_H(S)=N_G(S) \cap H = N_G(S)$, so the number of Sylow $p$-subgroups of $G$ and $H$ are equal. But $[G:S]=[G:H][H:N_G(S)][N_G(S):S]$ and taking the equation mod $p$ and using the fact that the number of Sylow $p$-subgroups $\equiv 1$ mod $p$ yields the required result.