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**JaneFairfax****Member**- Registered: 2007-02-23
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I think this is a nice result.

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**Ricky****Moderator**- Registered: 2005-12-04
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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?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**JaneFairfax****Member**- Registered: 2007-02-23
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The first mathematician to use the term group was Évariste Galois (according to the book *A Course in Group Theory* by John F. Humphreys). At about that same time, Augustin Louis Cauchy was independently studying these mathematical structures, but it was Galoiss terminology which became universally adopted.

http://z8.invisionfree.com/DYK/index.php?showtopic=118

*Last edited by JaneFairfax (2009-04-14 05:12:59)*

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**JaneFairfax****Member**- Registered: 2007-02-23
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**JaneFairfax****Member**- Registered: 2007-02-23
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*Last edited by JaneFairfax (2009-04-01 06:52:31)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

[align=center]

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

JaneFairfax wrote:

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Ricky wrote:

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.

*Last edited by JaneFairfax (2009-04-14 05:13:17)*

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