$gx=gy$ iff $x=y$ group action

Question

Let group $G$ act on set $X$, let $g \in G$ and $x,y \in X$. Is it true that $gx=gy$ iff $x=y$? I doubt that since group action is function $f(g,x)=gx$ and I think in order to be true function has to be injective. If not, is it true if action is double transitive? (Last question is because I read some pdf about double transitive group actions and author use $gx=gy$ iff $x=y$.

Answer 1

Let $x,y\in X$ and $g\in G$.

One direction is clear but subtle: group actions are functions by definition and are thus well-defined, meaning $x=y$ implies $g\cdot x=g\cdot y$.

We use this for the other direction. Suppose $g\cdot x=g\cdot y$. This gives

$$\begin{align} x&\stackrel{(1)}{=}e\cdot x\\ &=(g^{-1}g)\cdot x\\ &\stackrel{(2)}{=}g^{-1}\cdot(g\cdot x)\\ &\stackrel{(3)}{=}g^{-1}\cdot(g\cdot y)\\ &\stackrel{(2)}{=}(g^{-1}g)\cdot y\\ &=e\cdot y\\ &\stackrel{(1)}{=}y. \end{align}$$

Here the equalities marked $(1)$ hold by the identity property of a group action. Those marked by $(2)$ hold because of the associativity property of a group action. Finally, $(3)$ is true by well-definedness.