Index of smallest prime dividing $latex |G|$ implies Normal Subgroup

I have previously proved this at: Advanced Method for Proving Normal Subgroup. This is a neater, slightly shorter proof of the same theorem.

Index of smallest prime dividing $|G|$ implies Normal Subgroup
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:
(Hungerford pg 91)

Let $G$ act on the set $G/H$ (left cosets of $H$ in $G$) by left translation.

This induces a homomorphism $\sigma: G\to S_{G/H}\cong S_p$, where $\sigma_g(xH)=gxH$. Let $g\in\ker\sigma$. Then $gxH=xH$ for all $xH\in G/H$. In particular, when $x=1$, $gH=H$ which implies $g\in H$. So we have $\ker\sigma\subseteq H$.

Let $K=\ker\sigma$. By First Isomorphism Theorem, $G/K\cong\text{Im}\,\sigma\leq S_p$. Hence $|G/K|$ divides $|S_p|=p!$ But every divisor of $|G/K|=[G:K]$ must divide $|G|=|K|[G:K]$. Since no number smaller than $p$ (except 1) can divide $|G|$, we must have $|G/K|=p$ or $1$. However $\displaystyle |G/K|=[G:K]=[G:H][H:K]=p[H:K]\geq p.$

Therefore $|G/K|=p$ and $[H:K]=1$, hence $H=K$. But $K=\ker\sigma$ is normal in $G$.