Normalizers of Sylow p-subgroups 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?
 A: 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.
A: Here is another proof.  Let N denote the normalizer.  Let P be a sylow subgroup of G.
Thm: N(N(P)) subgroup N(P)
Pf:


*

*P sylow in G ==> P sylow in N(P)

*P sylow and normal in N(P) ==> P char in N(P)

*P char in N(P) and N(P) normal in N(N(P)) ==> P normal in N(N(P))

*P normal in N(N(P)) ==> N(N(P)) subgroup N(P)

A: I mentioned in the comments why your proof is incorrect.  Now here's a hint to get on track to finding the correct proof:
Hint: Note that $H$ is itself a group and $P$ is the unique (!) Sylow $p$-subgroup of $H$.  So if $g \in G$ normalizes $H$, where does conjugation by $g$ send $P$?
A: You want to prove that if $P$ is a Sylow subgroup of a finite group $G$ for some prime $p$, then $N_G(N_G(P)) = N_G(P)$.
Let $H = N_G(P)$, so $P \unlhd H$. Because $P$ is a Sylow subgroup of $G$, and $H \le G$, $P$ is also a Sylow subgroup of $H$. Because $P \unlhd H$, then $P$ is a unique Sylow subgroup of $H$ (ie $P$ is the only Sylow subgroup of $H$ of order $p$).
Let $g \in N_G(H)$, then:
$gPg^{-1} \le gHg^{-1} = H$.
Because every conjugate of a Sylow subgrooup is also a Sylow subgroup, then $gPg^{-1}$ is also a Sylow subgroup. Because $P$ is a unique Sylow subgroup of $H$, then $gPg^{-1} = P$. This means $g \in N_G(P) = H$.
So we have:

$g \in N_G(H)$
$g \in H$

We conclude $N_G(H) = H$ or $N_G(N_G(P)) = N_G(P)$ as required.
