Re: normal subgroup

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 07/20/04 

Date: Tue, 20 Jul 2004 08:12:19 -0500

On Tue, 20 Jul 2004 11:54:28 +0000 (UTC), afarzannia@yahoo.com (amineh
farzannia) wrote:

>I appreciate you for solving this problem .or please give me a hint.
>
>
>let G be a finite group . let H be asubgroup of G and let N be
>a normal subgroup of G .prove that if |H| and |G:N| are
>relatively prime . Then H is a subgroup of N .

Huh, I didn't know that...

There is a homomorphism f : H -> G/N defined by f(x) = xN.
Say K = f(H) is the image of H under this homomorphism.
Then K is a subgroup of G/N, which says something about
the order of K. But K is also isomorphic to H/ker(f),
which says something else about the order of K.
Hence the order of K is 1. This shows that H is a
subgroup of N, because ___.

************************

David C. Ullrich

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