A subgroup whose index is the smallest prime dividing $\abs{G}$ is normal

Let \(G\) act on the set \(G/H\) of left cosets by left translation, \(g\cdot xH=gxH\). This action is a homomorphism \(\varphi\colon G\to\operatorname{Sym}(G/H)\cong S_p\); let \(K\coloneqq\ker\varphi\). If \(g\in K\) then \(gH=H\), so \(K\subseteq H\), and \(K\normal G\). By the first isomorphism theorem \(G/K\) embeds in \(S_p\), so \([G:K]\) divides \(p!\). Also \([G:K]\) divides \(\abs{G}\), so every prime factor of \([G:K]\) is at least \(p\); the only such prime dividing \(p!\) is \(p\), and \(p^2\nmid p!\). Hence \([G:K]\in\theset{1,p}\). Since \(K\subseteq H\), \([G:K]\geq[G:H]=p\), so \([G:K]=p\) and \(K=H\). Therefore \(H=\ker\varphi\) is normal in \(G\).