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.

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- October 8th 2011, 10:09 AM #1

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## 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, 11:49 AM #2

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