basically it doesn't work because RIGHT multiplication is not a LEFT action:

(gh).a = agh

g.(h.a) = g.(ah) = ahg

if G is not abelian, then hg will be different than gh for at least one pair g,h.

Results 1 to 2 of 2

- October 8th 2011, 09:09 AM #1

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 7

## g\cdot a=ag

Let G be a group and let A = G. Show that if G is non-abelian then the maps defined by do not satisfy the axioms of a left group action of G on itself.

Since G is non-abelian, there exist .

Next it says this is a contradiction. Didn't the just show it worked? Is it a contradiction since this shows all elements in g commute? I am confused.

- October 8th 2011, 10:49 AM #2

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,445
- Thanks
- 788