Smallest-prime index forces normality.Suggest a slogan
Let \(G\) be a finite group and \(H \leq G\) a subgroup such that \(p \coloneqq [G:H]\) is the smallest prime dividing \(\abs{G}\). Then \(H \normal G\).
The action of \(G\) on the \(p\) left cosets of \(H\) by left translation gives a homomorphism \(\phi\colon G \to S_p\). Its kernel \(K\) satisfies \(K \subseteq H\), since \(gH = H\) for \(g \in K\), and \(K \normal G\). By the first isomorphism theorem, \(G/K\) embeds in \(S_p\), so \([G:K] = p\,[H:K]\) divides \(p!\), hence \([H:K]\) divides \((p-1)!\). Every prime divisor of \([H:K]\) divides \(\abs{G}\), so is at least \(p\), and every prime divisor of \((p-1)!\) is less than \(p\). Therefore \([H:K] = 1\), so \(H = K\) is normal in \(G\).