Let $G$ be group and $p$ be the least prime divisor of $|G|$.Then any subgroup of $G$ of index $p$ (if it exists) is normal in $G$.
My attempt $:$
Let $H \leq G$ be such that $[G:H] = p$.Let us consider the action of $G$ on the set $A$ of all left cosets of $H$ in $G$ (where the action is the left multiplication). Let $\pi_H$ denote the permutation representation of this action.Let $K = Ker\ \pi_H$. Then $K$ is a subgroup of $H$ (In fact $K$ is the largest normal subgroup of $G$ contained in $H$). Let $[H:K]=k$.Then $[G:K]=[G:H][H:K] = pk$. Since $\pi_H : G \longrightarrow Sym (A)$ is a homomorphism so by first isomorphism theorem we have $G/K \cong \pi_H (G)$. Since $|A| = [G:H] = p$. So $Sym (A) \cong S_p$. Hence $\pi_H (G)$ is a subgroup of $S_p$. This shows that $pk = |G/K| = |\pi_H (G)|$ divides $|S_p|=p!$ i.e. $k | (p-1)!$. Now since $k$ divides $|G|$ so if $k$ is prime or composite then by minimality of $p$, $k$ has to have a prime divisor $q \geq p$.Then $q | (p-1)!$ and since $q$ is prime so $q|(p-k)$ for some $1 \leq k \leq p-1$, a contradiction since $q \geq p$.Hence $k = 1$ and consequently $H = K$, completing the proof.
Is the above reasoning correct at all? Please verify it.
Thank you in advance.