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 $H\leq G$ with $\abs{G}$ finite and $\abs{G:H}=p$ , where $p$ is the smallest prime divisor of $\abs{G}$ , let $G$ act on the set $L$ of left cosets of $H$ in $G$ by left translation, and let $\varphi:G\rightarrow S_p$ be the homomorphism induced by this action. Now, if $g\in\ker\varphi$ , then $gxH=xH$ for each $x\in G$ , and in particular, $gH=H$ , whence $g\in H$ . Thus $K=\ker\varphi$ 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 $\abs{G/K}=\abs{G:K}$ must divide $p!$ ; moreover, any divisor of $\abs{G:K}$ must also divide $\abs{G}=\abs{G:K}\abs{K}$ , and because $p$ is the smallest divisor of $\abs{G}$ different from $1$ , the only possibilities are $\abs{G:K}=p$ or $\abs{G:K}=1$ . But $\abs{G:K}=\abs{G:H}\abs{H:K}=p\abs{H:K}\geq p$ , which forces $\abs{G:K}=p$ , and consequently $\abs{H:K}=1$ , so that $H=K$ , from which it follows that $H$ is normal in $G$ .