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This question already has an answer here:

This is a problem from Herstein that I have been stuck upon for ages. I am becoming increasingly disappointed and disillusioned about my abilities due to this problem.

Let G be finite and let $p$ be the smallest prime dividing $|G|$. Let $H \le G$ be of index $p$. Prove that $H$ is a normal subgroup of $G$.

I am nowhere near getting a solution. Assume the statement is not true, so $N(H) \ne G$ then $H \le N(H) \le G$ implies that $N(H)=H$. How can I get a contradiction from this? I am trying to find a homomorphism which has H has its kernel, but getting nowhere. I would like to see, atleast a couple of different solutions for this.

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merged by robjohn Feb 10 '13 at 12:25

This question was merged with Normal subgroup of prime index because it is an exact duplicate of that question.