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 implies Normal Subgroup
If is a subgroup of a finite group of index , where is the smallest prime dividing the order of , then is normal in .
(Hungerford pg 91)
Let act on the set (left cosets of in ) by left translation.
This induces a homomorphism , where . Let . Then for all . In particular, when , which implies . So we have .
Let . By First Isomorphism Theorem, . Hence divides But every divisor of must divide . Since no number smaller than (except 1) can divide , we must have or . However
Therefore and , hence . But is normal in .