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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. smile


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

Ricky
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Registered: 2005-12-04
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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
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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. smile

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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Registered: 2007-02-23
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Re: Group theory


[align=center]

[/align]




smile


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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


smile

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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Registered: 2007-02-23
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Re: Group theory

[align=center]

[/align]


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

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

smile


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

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

smile


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

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

JaneFairfax wrote:

   

smile


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

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



smile


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

JaneFairfax
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Registered: 2007-02-23
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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. Cheering.gif

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. wink

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


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