normality of subgroups of prime index

Proposition.

If $H$ is a subgroup of a finite group $G$ of index $p$, where $p$ is the smallest prime dividing the order of $G$, then $H$ is normal in $G$.

Proof.

Suppose $H\le G$ with $|G|$ finite and $|G:H|=p$, where $p$ is the smallest prime divisor of $|G|$, let $G$ act on the set $L$ of left cosets of $H$ in $G$ by left , and let $\phi :G\to S_p$ be the http://planetmath.org/node/3820homomorphism induced by this action. Now, if $g\in \ker \phi$, then $gxH=xH$ for each $x\in G$, and in particular, $gH=H$, whence $g\in H$. Thus $K=\ker \phi$ is a normal subgroup of $H$ (being contained in $H$ and normal in $G$). By the First Isomorphism Theorem, $G/K$ is isomorphic to a subgroup of $S_p$, and consequently $|G/K|=|G:K|$ must http://planetmath.org/node/923divide $p!$; moreover, any divisor of $|G:K|$ must also $|G|=|G:K||K|$, and because $p$ is the smallest divisor of $|G|$ different from $1$, the only possibilities are $|G:K|=p$ or $|G:K|=1$. But $|G:K|=|G:H||H:K|=p|H:K|\ge p$, which $|G:K|=p$, and consequently $|H:K|=1$, so that $H=K$, from which it follows that $H$ is normal in $G$.∎