Generalizing the case p=2 we would like to know if the statement below is true.

Let p the smallest prime dividing the order of G. If H is a subgroup of G with index p then H is normal.

**Answer**

This is a standard exercise, and the answer is that the statement is true, but the proof is rather different from the elementary way in which the p=2 case can be proven.

Let H be a subgroup of index p where p is the smallest prime that divides |G|. Then G acts on the set of left cosets of H, {gH∣g∈G} by left multiplication, x⋅(gH)=xgH.

This action induces a homomorphism from G→Sp, whose kernel is contained in H. Let K be the kernel. Then G/K is isomorphic to a subgroup of Sp, and so has order dividing p!. But it must also have order dividing |G|, and since p is the smallest prime that divides |G|, it follows that |G/K|=p. Since |G/K|=[G:K]=[G:H][H:K]=p[H:K], it follows that [H:K]=1, so K=H. Since K is normal, H was in fact normal.

**Attribution***Source : Link , Question Author : Sigur , Answer Author : Arturo Magidin*