Let $G$ be a group and let $S=G.$ Prove that the following $g\cdot x∶=gxg^{−1}$ for all $g \in G$ and $x \in S$ is an action of $G$ on $S$.

Question

I am proving that it is associative first, but this is what I need help with:

Let $g,h \in G.$ First we show that it is associative. Then $g \cdot(h \cdot x)= g \cdot (hxh^{-1})= g(hxh^{-1})g^{-1}.$

I am wondering if the next step would be to say $(gh)xh^{-1}g^{-1}= (gh)x(gh)^{-1}=(gh) \cdot x.$ I donʻt even know if this is correct, also my end result doesnʻt have the group action in it. To show its associative it would be to show that $ g \cdot(h \cdot x)= (g \cdot h)\cdot x$ right?

Answer 1

You got

$$\begin{align} g\cdot (h\cdot x)&=[ \dots ]\\ &=(gh)xh^{-1}g^{-1}\\ &= (gh)x(gh)^{-1}\\ &=(gh) \cdot x \end{align}$$ right.

What remains is to show that $e\cdot x=x$. Indeed, we have $e\cdot x=exe^{-1}=x$ (since $e^{-1}=e$).