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.

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 $P$.
