Indeed, this is a rather cool result. In fact, it generalizes fairly well. That is, if G contains a subgroup H of index p for the smallest prime dividing the order of G, then H is normal in G.

Do you know how to prove this?

Ive just read it *A Course in Group Theory* by John F. Humphreys! It is stated as Corollary 9.25 on page 86.

I didnt know how to prove it, so I quote Humphreyss proof.

Apply Corollary 9.23 to the subgroup

Hto find a normal subgroupNwith |G:N| dividingp! Since |G:N| also divides |G|, it divides the greatest common divisor ofp! and |G|. Sincepis the smallest prime dividing |G|, the greatest common divisor ofp! and |G|, so |G:N| =p= |G:H|. SinceNis contained inH, it follows thatNis equal toH.

I have read every proof in my book carefully, and have followed every proof so far. I can thus safely say that I havent missed any trick, big or small.

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Do you know how to prove this?

]]>I think this is a nice result.

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