# 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.

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The number of Sylow 7-subgroups is congruent to 1 mod 7, so is either 1 (when the Sylow subgroup is normal) or 8 (other possibilities are easy to dismiss). G acts on the Sylow 7-subgroups by conjugation, and the normaliser N of P is the subgroup of G which sends P to itself. This action is transitive on the 8 Sylow 7-subgroups so N has order 168/orbit of P = 168/8=21. There are limited options now - so consider those in turn. – Mark Bennet Aug 13 '12 at 20:10

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 $P$.
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 $[G:N] = [G:M][M:N] \geq (p+1)^{2}.$