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

• May 26th 2010, 07:09 PM
davismj
[SOLVED] Normalizer and Equality of left, right cosets.
http://i50.tinypic.com/33o7hp2.jpg

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 $\displaystyle h \in H, aha^{-1} = h' \in H$ with $\displaystyle h' \neq h$ isn't a possibility.
• May 26th 2010, 07:46 PM
Drexel28
What's the huge assumption?
• May 26th 2010, 08:06 PM
davismj
Quote:

Originally Posted by Drexel28
What's the huge assumption?

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

Edit: Maybe thats not a huge assumption since

$\displaystyle aHa^{-1} = H \implies aHa^{-1}(a) = H(a) \implies aH = Ha$.
• May 26th 2010, 08:12 PM
Drexel28
Quote:

Originally Posted by davismj
$\displaystyle aHa^{-1} = H \implies aH = Ha$

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

If $\displaystyle ah\in aH$ then $\displaystyle aha^{-1}\in aHa^{-1}=H$ and thus $\displaystyle aha^{-1}=h'$ for some $\displaystyle h'\in H$ and thus $\displaystyle ah=h'a$ in particular $\displaystyle ah\in Ha$ so that $\displaystyle aH\subseteq Ha$. The reverse inclusion is similar.
• May 26th 2010, 08:23 PM
davismj
Quote:

Originally Posted by Drexel28
$\displaystyle aHa^{-1}=H\implies aH=Ha$?

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

Thanks bud.