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JaneFairfax

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Registered: 2007-02-23

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Group theory

I think this is a nice result.

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?

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"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..."

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.

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

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

JaneFairfax

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Re: Group theory

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JaneFairfax

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Re: Group theory

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

JaneFairfax

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Registered: 2007-02-23

Posts: 6,868

Re: Group theory

[align=center]

[/align]

JaneFairfax

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Registered: 2007-02-23

Posts: 6,868

Re: Group theory

JaneFairfax

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Registered: 2007-02-23

Posts: 6,868

Re: Group theory

JaneFairfax

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Registered: 2007-02-23

Posts: 6,868

Re: Group theory

   

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Group theory

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

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?

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)