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## #1 2009-04-24 09:16:54

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

### The JordanHölder theorem

I gather the author wants to prove it in three stages. The first stage is to establish something called the Zassenhaus lemma. The proof is not all that difficult to follow, but involves a lot of tricky details  I was stuck on a particularly tricky bit for ages before I finally realized what was going on. The second stage is Shreiers refinement theorem  whose proof using the Zassenhaus lemma is a total piece of cake. And the third stage, I suppose, will be to prove the JordanHölder theorem using Shreiers theorem.

Last edited by JaneFairfax (2009-04-24 09:17:43)

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## #2 2009-04-26 04:38:53

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

### Re: The JordanHölder theorem

The Zassenhaus lemma is merely a result about normal subgroups.

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The way to prove this is to prove that

and that the LHS is isomorphic to
; it will follow by symmetry that the RHS is also isomorphic to this group and hence to the LHS.

According to Wikipedia, this result is also called the butterfly lemma. There is a lovely picture of a butterfly on the Wikipedia article as well.

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