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## #1 2008-06-27 11:39:51

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

### Group theory

I think this is a nice result.

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## #2 2008-10-08 10:10:55

Ricky
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Registered: 2005-12-04
Posts: 3,791

### Re: Group theory

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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## #3 2009-03-27 22:37:40

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

### Re: Group theory

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 Galoiss 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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## #4 2009-04-01 01:01:45

JaneFairfax
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[align=center]

[/align]

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## #5 2009-04-01 01:31:34

JaneFairfax
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Registered: 2007-02-23
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### Re: Group theory

Last edited by JaneFairfax (2009-04-01 06:52:31)

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## #6 2009-04-01 01:35:42

JaneFairfax
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[align=center]

[/align]

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## #7 2009-04-06 12:18:33

JaneFairfax
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## #8 2009-04-08 04:42:23

JaneFairfax
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## #9 2009-04-10 02:20:28

JaneFairfax
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### Re: Group theory

JaneFairfax wrote:

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## #10 2009-04-10 02:32:19

JaneFairfax
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## #11 2009-04-12 02:27:06

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

### Re: Group theory

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?

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

I didnt know how to prove it, so I quote Humphreyss 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.

states that there is a normal subgroup
with
and
. In fact,
is the kernel of a homomorphism mapping
to the permutation group of the set of all left cosets of
in
.

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

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

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