= Subgroup of index the smallest prime divisor is normal
{title2=$[G:H]=p=\min\{q:q\text{ prime},\ q\mid|G|\}\ \Longrightarrow\ H\trianglelefteq G$}
Let $G$ be finite and let $H$ have index $p$, the smallest prime divisor of $|G|$. The <group action> on <left cosets> gives $\Phi:G\to S_p$, with $\ker\Phi\subseteq H$. The image order divides both $|G|$ and $p!$. By the choice of $p$, the only possible nontrivial prime factor common to these orders is $p$, and $p!$ contains only one factor of $p$. Thus the image order is $1$ or $p$. The action is <transitive> on $p>1$ cosets, so its image is nontrivial and has order $p$. The <first isomorphism theorem> gives $|\ker\Phi|=|G|/p=|H|$, hence $\ker\Phi=H$. A <kernel of a group homomorphism> is a <normal subgroup>, proving the claim. For example, every subgroup of order $25$ in a group of order $75$ is normal.
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