Dummit and Foote define a group action as follows:

A group action of a group G on a set A is a map from G x A to A (written as g$\displaystyle \cdot$a, for all g$\displaystyle \in$ G and a$\displaystyle \in$A) satisfying the following properties

(1) $\displaystyle g_1$$\displaystyle \cdot$($\displaystyle g_2$$\displaystyle \cdot$a) = ($\displaystyle g_1$$\displaystyle g_2$)$\displaystyle \cdot$a for all $\displaystyle g_1$, $\displaystyle g_2$$\displaystyle \in$G and all a$\displaystyle \in$A

(2) 1$\displaystyle \cdot$a = a for all a$\displaystyle \in$A

Dummit and Foote Section 1.7 Group Actions Exercise 14 reads as follows:

Let G be a group and let A = G. SHow that if G is non-abelian then the maps defined by g$\displaystyle \cdot$a = ag for all g, a $\displaystyle \in$ G dosatisy the axioms of a (left) group action of G on itselfnot

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I am unsure of whether the operation of the group G on the set G - that is g$\displaystyle \cdot$a can be treated the same as the operation of two elements of G as a group - that is ga.

Can anyone clarify this? [I suspect it cannot!]

Anyway, proceeding withthe problem we need to show

(1) $\displaystyle g_1$$\displaystyle \cdot$($\displaystyle g_2$$\displaystyle \cdot$a) = ($\displaystyle g_1$$\displaystyle g_2$)$\displaystyle \cdot$a for all $\displaystyle g_1$, $\displaystyle g_2$$\displaystyle \in$G and all a$\displaystyle \in$A is NOT satisfied

Then $\displaystyle g_1$$\displaystyle \cdot$($\displaystyle g_2$$\displaystyle \cdot$a) = $\displaystyle g_1$$\displaystyle \cdot$(a$\displaystyle g_2$)

The term on the right above is strictly a group action operation of $\displaystyle g_1$ on a set element a$\displaystyle g_2$

How do I progress this?

I do not think I can write ($\displaystyle g_1$a) $\displaystyle g_2$ as there is a group action mixed with a group operation ???

Am I overthinking this?

Peter