TITLE: Normal subgroup of prime index
QUESTION [102 upvotes]: Generalizing the case $p=2$ we would like to know if the statement below is true.
Let $p$ the smallest prime dividing the order of $G$. If $H$ is a subgroup of $G$ with index $p$ then $H$ is normal.

REPLY [6 votes]: Another answer if you know character theory over the complex numbers. Let $H$ be a subgroup of order $G$ of index $p$, smallest prime dividing $|G|$. Look at the trivial character of $H$ and induce it to $G$. Then $1_H^G=\sum_{\chi \in{\rm Irr}(G)} a_{\chi}\chi$, with $a_{\chi} \in \mathbb{Z}_{\geq 0}$. Since $[1_H^G, 1_G]=[1_H,1_H]=1=a_{1_G}$, it follows that the irreducible constituents of $1_H^G$ have degree $\chi(1) \leq p-1 \lt p$. Since these degrees divide $|G|$ is follows that all the irreducible constituents $\chi$ must be linear, that is $\chi(1)=1$. Hence $H \supset \ker(1_H^G)=\cap\{{\ker(\chi) \in{\rm Irr}(G): a_{\chi} \gt 0}\} \supseteq G'$, whence $H \lhd G$.