# Abstract Algebra proof - cosets

• Apr 27th 2009, 05:57 PM
kathrynmath
Abstract Algebra proof - cosets
Let H be a subgroup of G such that g^-1hg is an element of H for all g in G and all h in H. Show that every left coset gH is the same as the right coset Hg.

need to show gh1=h2g
I know I need to show this, but am unsure on how to.
I know the whole set up excpet for this part and it confuses me.
• Apr 27th 2009, 06:21 PM
rtblue
maybe i'm just really bad at math, but... isn't this question in the wrong forum area?? just a though. After all.. it is abstract algebra, and i think MHF has a different forum section for those types of questions.
• Apr 28th 2009, 02:07 AM
pickslides
This should be in:

Universtiy Math Help > Number Theory
• Apr 28th 2009, 02:51 AM
pickslides
Quote:

Originally Posted by kathrynmath
Let H be a subgroup of G such that g^{-1}hg is an element of H for all g in G and all h in H. Show that every left coset gH is the same as the right coset Hg.

need to show gh1=h2g
I know I need to show this, but am unsure on how to.
I know the whole set up excpet for this part and it confuses me.

I'm a little rusty on the group theory stuf but I can recall the following:

Every subgroup of an abelian group is normal. As your subgroup H is normal as $\displaystyle g^-1hg \in H$ maybe we can assume G is abelian (help anyone!)

So if you can prove that G is in fact abelian then the following should follow for your cosets

if $\displaystyle H \triangleleft G$ then

$\displaystyle g_1Hg_2H$

operating on cosets

$\displaystyle \Rightarrow g_1g_2H$
$\displaystyle \Rightarrow g_2g_1H$

I hope this helps some.