Solution
= Solution
{c}
Write $g\cdot x=gxg^{-1}$. If $e$ is the <identity element> of $G$, then $e\cdot x=exe^{-1}=x$. Also, for $g,h,x\in G$,
$$
g\cdot(h\cdot x)=g(hxh^{-1})g^{-1}
=(gh)x(h^{-1}g^{-1})=(gh)x(gh)^{-1}
=(gh)\cdot x.
$$
The middle equality uses <associativity> and the <inverse element> rule $(gh)^{-1}=h^{-1}g^{-1}$. Thus \b[this is a left <group action>], the <conjugation action> of $G$ on itself.