A subgroup whose index is the smallest prime divisor of the order is normal

Statement

Let $G$ be a finite group and let $p$ be the smallest prime number that divides $|G|$. Let $H$ be a subgroup of $G$ of index $p$. Then $H$ is normal in $G$.

Remarks

For $p = 2$ this is the fact that subgroups of index two are normal. The proof lets $G$ act on the $p$ left cosets of $H$. The kernel $K$ of this action lies in $H$, and $[G : K]$ divides both $|G|$ and $p!$. Since $p$ is the smallest prime divisor of $|G|$, this forces $[G : K] = p$, so $H = K$ is normal.

Explanations

L1 human

Index $2$ is not the only small index that forces normality. If the index is the smallest prime that divides the order of the group, the subgroup is also normal. For example, a subgroup of index $3$ in a group of order $75 = 3 \cdot 5^2$ is normal.