Sylow 7-subgroups in a group of order 168 Another question from my qual studying that's been stumping me.  I'm still a little lost on normalizers.  The question is:
Let G be a group of order 168, and let P be a Sylow 7-subgroup of G.  Show that either P is a normal subgroup of G or else the normalizer of P is a maximal subgroup of G.
 A: A way to generalize is this: Let $G$ be a finite group and $P$ be a Sylow $p$-subgroup of $G$. Let $N= N_{G}(P).$ Then if $1 < [G:N] < (p+1)^{2},$ the subgroup $N$ must be maximal. For suppose that $M$ is a proper subgroup of $G$ which strictly contains $N.$ Then we have $[M:N] \equiv 1$ (mod $p$) by Sylow's theorem, since $N = N_{M}(P)$ and $P$ is still a Sylow $p$-subgroup of $M.$ Also, since $[G:N] \equiv 1 $ (mod $p$) and $[M:N] \equiv 1$ (mod $p$), we must have $[G:M] \equiv 1$ (mod $p$). Since $[G:M] >1$ and $[M:N] >1$ by assumption, we have $[G:N] = [G:M][M:N] \geq (p+1)^{2}.$
A: Suppose $P$ is not normal. Then by Sylow's theorems, $G$ has $8$ $7$-Sylows, all of them conjugate to $P$. Thus the normalizer $N$ of $P$ has index $8$ in $G$, and hence order $21$. Suppose $N$ is not maximal, so there's a subgroup $H$ of $G$ strictly between $N$ and $G$. Then $H$ must have order $42$ or $84$. Sylow's theorems show that $H$ has a normal $7$-Sylow, which must be $P$. But then $P$ is normal in $H$, contradicting the fact that $H$ properly contains the normalizer $N$ of $P$.
