# Normal subgroup of prime index

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.

This is a standard exercise, and the answer is that the statement is true, but the proof is rather different from the elementary way in which the $$p=2p=2$$ case can be proven.
Let $$HH$$ be a subgroup of index $$pp$$ where $$pp$$ is the smallest prime that divides $$|G||G|$$. Then $$GG$$ acts on the set of left cosets of $$HH$$, $${gH∣g∈G}\{gH\mid g\in G\}$$ by left multiplication, $$x⋅(gH)=xgHx\cdot(gH) = xgH$$.
This action induces a homomorphism from $$G→SpG\to S_p$$, whose kernel is contained in $$HH$$. Let $$KK$$ be the kernel. Then $$G/KG/K$$ is isomorphic to a subgroup of $$SpS_p$$, and so has order dividing $$p!p!$$. But it must also have order dividing $$|G||G|$$, and since $$pp$$ is the smallest prime that divides $$|G||G|$$, it follows that $$|G/K|=p|G/K|=p$$. Since $$|G/K|=[G:K]=[G:H][H:K]=p[H:K]|G/K| = [G:K]=[G:H][H:K] = p[H:K]$$, it follows that $$[H:K]=1[H:K]=1$$, so $$K=HK=H$$. Since $$KK$$ is normal, $$HH$$ was in fact normal.