Let \(G\) be a finite group, and let \(H \le G\) be a subgroup of index \([G : H] = p\), where \(p\) is the smallest prime dividing \(|G|\). Prove that \(H\) is a normal subgroup of \(G\) (\(H \trianglelefteq G\)).
Solution
Let \(G\) act on the \(p\) left cosets of \(H\). This gives \[ \rho:G\to S_p, \] and let \(K=\ker\rho\). Then \(K\trianglelefteq G\), \(K\le H\), and \[ [G:K]=|\operatorname{im}\rho|\mid p!. \] Since \([G:H]=p\), \[ [G:K]=p[H:K], \] so \([H:K]\mid (p-1)!\).
On the other hand, \([H:K]\mid |G|\). If \([H:K]>1\), choose a prime \(q\mid[H:K]\). Then \(q\mid|G|\), so by minimality of \(p\) we have \(q\ge p\). But \(q\mid(p-1)!\), so \(q<p\), a contradiction. Hence \([H:K]=1\), so \(H=K\trianglelefteq G\).