Let $G$ be a finite group, and let $H$ be a subgroup of $G$ such that the index $(G\colon H)$ of $H$ in $G$ is the smallest prime that divides the order of $G$. Can we say anything about whether or not $H$ is normal in $G$?
The answer is of course yes if $G$ is abelian.
Denote set of left cosets by $G/H$ and let $S\left(G/H\right)$ denote the symmetric group of bijections on it.
Then $f:G\rightarrow S\left(G/H\right)$ defined by $g\mapsto\lambda_{g}$, where $\lambda_{g}\in S\left(G/H\right)$ on its turn is defined by $xH\mapsto gxH$, is a grouphomomophism with kernel $N:=\cap_{x\in G}xHx^{-1}$.
Here $S\left(G/H\right)$ and $S_{p}$ are isomorphic and $G/N\simeq f\left(G\right)\leq S\left(G/H\right)$ so $\left[G:N\right]$ divides $p!$.
We have $N\subset H\subset G$ with $\left[G:N\right]=p\left[H:N\right]$ so that $\left[H:N\right]$ divides $\left(p-1\right)!$ and $\left|G\right|$.
But $\left|G\right|$ and $\left(p-1\right)!$ are coprime leading to $\left[H:N\right]=1$ or equivalently $H=N$.
As kernel of $f$ subgroup $H$ is normal.