g\dot a=ag

**dwsmith** · October 8th 2011, 09:09 AM

Let G be a group and let A = G. Show that if G is non-abelian then the maps defined by $g\cdot a=ag, \ \forall g,a\in G$ do not satisfy the axioms of a left group action of G on itself.

Since G is non-abelian, there exist $g_1,g_2\in G \ \text{s.t} \ g_1g_2\neq g_2g_1$ .

$eg_1g_2=a^{-1}ag_1g_2=a^{-1}((g_1g_2)\cdot a)=a^{-1}(g_1(g_2\cdot a))=a^{-1}(g_1\cdot (ag_2))=a^{-1}(ag_2g_1)=g_2g_1$

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.

**Deveno** · October 8th 2011, 10:49 AM

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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