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?

  • $\begingroup$ Are the operations in $g\circ x \mapsto x * g^{-1}$, $\circ$ and $*$ the same? $\endgroup$ Jun 23, 2020 at 8:30
  • $\begingroup$ sorry the operation should be: $ (g,x) \mapsto x*g^{-1} $ $\endgroup$
    – bigbuddies
    Jun 23, 2020 at 8:33
  • 3
    $\begingroup$ HInt: $(gh)^{-1}=h^{-1}g^{-1}$. $\endgroup$ Jun 23, 2020 at 8:36
  • $\begingroup$ With reference to left-cancellation law, I state that this left-action is a property of an element, that is in the group $\endgroup$ Jun 23, 2020 at 8:37
  • $\begingroup$ If I am understanding your question correctly, thenthe answer is simple. It is a left group action because it is a group action in which the $g$ is on the left of the $x$. A left group action of a group $G$ on a set $X$ is a map $G \times X \to X$, and a right group action is a map $X \times G \to X$. $\endgroup$
    – Derek Holt
    Jun 23, 2020 at 9:27

3 Answers 3


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.


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}


enter image description here

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

enter image description here

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

enter image description here

  • $\begingroup$ Hm I think I am making it a bit more difficult than it should be, but is it something like this: $(g,x)=(h,x) \implies xg^{-1}=xh^{-1} \implies g^{-1}=h^{-1} \implies g=h \implies gx=hx$ ? $\endgroup$
    – bigbuddies
    Jun 23, 2020 at 9:09
  • $\begingroup$ No, It cannot be approached like the left cancellation proof. The more i read the definition of the left group action, the more I agree on this. $\endgroup$ Jun 23, 2020 at 9:16
  • $\begingroup$ @Josephine What are the sets G and S given to you? $\endgroup$ Jun 23, 2020 at 9:22
  • $\begingroup$ what do you mean? I just know that G is a group and X is a general collection of things, nothing specific. I guess they just want me to think about the definition of left group actions and make me realize that $ gx:= xg^{-1} $, but I am having a hard time seeing it $\endgroup$
    – bigbuddies
    Jun 23, 2020 at 9:25
  • $\begingroup$ @Josephine don't worry, this "seeing" need not be visual seeing like looking at the mountains but axiomatic-application seeing like a disproving reasoning. Carrying on, so G is a group and S (here X) is a set and the mapping(rule) is that $gx:= xg^{-1}$. $\endgroup$ Jun 23, 2020 at 9:33

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