How do i show any subgroup of index $p$ is normal?

Question

If $G$ is a finite group of order $n$ and $p$ is the smallest prime dividing $|G|$, then prove that any subgroup of index $p$ is normal.

This falls under the category of Cayley's Theorem, but i have difficulty in proving this. Please help me out!

Answer 1

$G$ acts by left multiplication on the cosets $G/H$, giving rise to a homomorphism $G\to S_p$. The restriction to $H$ leaves the coset $eH$ fix, hence is in fact mapped into $S_{p-1}$. From the minimality of $p$, conclude that $H\to S_{p-1}$ must be trivial.

Answer 2

Suppose $H \leq G, |G:H|=p$. Let $\pi_H$ be the permutation representation afforded by multiplication on the set of left cosets of $H \text{in} G,$ and let $K=\text{ker}_{\pi_H}$ and let $|H:K|=k$. Then, $|G:K|=|G:H||H:K|=pk$. Now since $H$ has $p$ left cosets, $G/K \cong$ to a subgroup of $S_p$, the symmetric group of order $p!$, by the First Isomorphism Theorem. By Lagrange's Theorem, $pk=|G/K|$ divides $p! \implies k|(p-1)!$, but all prime divisors of $(p-1)!$ are less than $p$ and by the minimality of $p$, every divisor of $k$ is greater than or equal to $p$. And this forces $k=1$, so $H=K \mathrel{\unlhd} G$.