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

Let us write an action as a function $A:G\times X \to X$.
Then, for $A$ to be a left action, it has to satisfy $A(g, A(h, x)) = A(gh, x)$.
In contrast, right actions satisfy $A(g, A(h,x)) = A(hg, x)$.
(In fact, right actions are usually written as $B:X\times G\to X$ satisfying $H(H(x, g), h) = H(x, gh)$ to make it looks more like an associativity law.

So for your question, you need to show:

$A(g, A(h,x)) = A(g, x*h^{-1}) = x*h^{-1}*g^{-1} = x*(gh)^{-1} = A(gh, x)$
So even if the $g$ 'appears' on the right, it is indeed a left action.

Answer 2

Here, our mapping takes a ordered set from the cartesian product $G\ * \ X$ to an element in set $X$, $(g,x):=xg^{-1}$

If we let the mapping $(g,x)\mapsto x*g^{-1}$ be a left group action of G on S.

Then, by that very assumption, we can now say that the mapping must satisfy the axioms of the left group actions.

P.S. If your mapping satisfies the axioms of the left group actions, it implies that

Answer 3

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}