[SOLVED] Normalizer and Equality of left, right cosets.

**davismj** · May 26th 2010, 07:09 PM

I think this is a huge assumption to make, but I think this is true. I'm not sure how to bridge the gap.

I'm still kind of confused why for some group H and some $h \in H, aha^{-1} = h' \in H$ with $h' \neq h$ isn't a possibility.

**Drexel28** · May 26th 2010, 07:46 PM

What's the huge assumption?

**davismj** · May 26th 2010, 08:06 PM

Originally Posted by Drexel28

What's the huge assumption?

$aHa^{-1} = H \implies aH = Ha$

Edit: Maybe thats not a huge assumption since

$aHa^{-1} = H \implies aHa^{-1}(a) = H(a) \implies aH = Ha$ .

**Drexel28** · May 26th 2010, 08:12 PM

Originally Posted by davismj

$aHa^{-1} = H \implies aH = Ha$

$aHa^{-1}=H\implies aH=Ha$ ?

If $ah\in aH$ then $aha^{-1}\in aHa^{-1}=H$ and thus $aha^{-1}=h'$ for some $h'\in H$ and thus $ah=h'a$ in particular $ah\in Ha$ so that $aH\subseteq Ha$ . The reverse inclusion is similar.

**davismj** · May 26th 2010, 08:23 PM

Originally Posted by Drexel28

$aHa^{-1}=H\implies aH=Ha$ ?

If $ah\in aH$ then $aha^{-1}\in aHa^{-1}=H$ and thus $aha^{-1}=h'$ for some $h'\in H$ and thus $ah=h'a$ in particular $ah\in Ha$ so that $aH\subseteq Ha$ . The reverse inclusion is similar.

Thanks bud.

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[SOLVED] Normalizer and Equality of left, right cosets.

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