If $|G:H|=p$ where $p$ is the smallest prime dividing $|G|$ then $H\unlhd G$.
I have come up with a proof as follows, which may be probably wrong (since my proof has no used the condition that $p$ is the smallest prime dividing $|G|$), but I can not find where the mistake is. Could anyone help me have a look?
Consider the permutation representation $\pi_H:G\to\mathrm{S}_{G/H}$ where $G/H$ denotes the set of left cosets of $H$. It has kernel $K:=\cap_{x\in G}xHx^{-1}$ which is contained in $H$. Since $K\unlhd G$, $|G/K:H/K|=|G:H|=p$. And since $|G/K|<|G|$, by induction we have $H/K\unlhd G/K$, then by the Fourth Isomorphism Theorem, $H\unlhd G$.
Edit: I have found where the mistake was: the quotient $G/K$ is not necessarily of order strictly smaller then the one of $G$. For instance, if $H<G$ is proper and $G$ is simple, then $K\leqslant H<G$ implies $K=1$ and hence $|G/K|=|G|\not<|G|$, so we could not use induction to deduce that $H/K\unlhd G/K$.
