Index of smallest prime dividing $latex |G|$ implies Normal Subgroup

I have previously proved this at: Advanced Method for Proving Normal Subgroup. This is a neater, slightly shorter proof of the same theorem.

Index of smallest prime dividing |G| implies Normal Subgroup
If H is a subgroup of a finite group G of index p, where p is the smallest prime dividing the order of G, then H is normal in G.

Proof:
(Hungerford pg 91)

Let G act on the set G/H (left cosets of H in G) by left translation.

This induces a homomorphism \sigma: G\to S_{G/H}\cong S_p, where \sigma_g(xH)=gxH. Let g\in\ker\sigma. Then gxH=xH for all xH\in G/H. In particular, when x=1, gH=H which implies g\in H. So we have \ker\sigma\subseteq H.

Let K=\ker\sigma. By First Isomorphism Theorem, G/K\cong\text{Im}\,\sigma\leq S_p. Hence |G/K| divides |S_p|=p! But every divisor of |G/K|=[G:K] must divide |G|=|K|[G:K]. Since no number smaller than p (except 1) can divide |G|, we must have |G/K|=p or 1. However \displaystyle |G/K|=[G:K]=[G:H][H:K]=p[H:K]\geq p.

Therefore |G/K|=p and [H:K]=1, hence H=K. But K=\ker\sigma is normal in G.

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