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 H to find a normal subgroup N with |G : N| dividing p! Since |G : N| also divides |G|, it divides the greatest common divisor of p! and |G|. Since p is the smallest prime dividing |G|, the greatest common divisor of p! and |G|, so |G : N| = p = |G : H|. Since N is contained in H, it follows that N is equal to H.
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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[/align]Do you know how to prove this?
]]>I think this is a nice result.
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