# Normalizer subgroup proof proving the inverse

1. Sep 17, 2009

### cmj1988

This specifically relates more towards the argument as to why an inverse exists.

First the problem

The normalizer is defined as follows, NG(H)={g-1Hg=H} for some g in NG(H). I get why identity exists and why the operation is closed. It is in arguing that an inverse exists that I have beef. Specifically this argument:

eHe= (g-1)-1g-1Hgg-1=(g-1)-1Hg-1

so g-1 is in NG(H)

This above proof I found in the book Groups, rings, and fields by D.A.R. Wallace.

I wondering why this is more valid than this:

H=gHg-1
g-1H=Hg-1
g-1Hg=H

So g-1 is in NG(H)

I'm sorry if this question is so dense someone has an aneurysm

Last edited: Sep 17, 2009
2. Sep 18, 2009

### cmj1988

I have another proof, but I'm not sure why this is any more valid. It follows the same route as the first one I presented up to this point:

gHg^{-1}
This is equivalent to
(g^{-1})^{-1}Hg^{-1}

Therefore g^{-1} is in N_G(H)