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 .

**Proof:**

(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 .