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

**davismj** · May 26th 2010, 08:09 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, 08:46 PM

What's the huge assumption?

**davismj** · May 26th 2010, 09:06 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, 09:12 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, 09:23 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.

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

LinkBack

Thread Tools

Display

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

Similar Math Help Forum Discussions

Left and Right cosets.

Left cosets

left and right cosets

Left and Right Cosets

right and left cosets

Search Tags