Proof $(g,x) \mapsto x * g^{-1}$ is a left group action.

Question

I am getting a bit confused with group actions. I have to prove that the map $(g,x)\mapsto x*g^{-1}$ is a left group action. I have already proven it to be a group action using the fact that it is associative, and there is an identity element. How do I go about proving that it is a left group action? Any ideas?

Answer 1

If $x*g$ is a (right) action ("$hyp.$"), then $g\cdot x := x*g^{-1}$ is a left action. In fact:

1. $\space\space e\cdot x=x*e^{-1}=x*e\stackrel{hyp.}{=}x, \space\forall x\in X$;

2. \begin{alignat}{1} (gh)\cdot x &= x * (gh)^{-1} \\ &= x*(h^{-1}g^{-1}) \\ &\stackrel{hyp.}{=} (x*h^{-1})*g^{-1} \\ &= g\cdot (x*h^{-1}) \\ &= g\cdot (h\cdot x) \\ \end{alignat}