If $G$ acts on $X$ then $\psi: X\to X$ is a bijection

Question

I would appreciate if you could please express your opinion on this proof. I don't know how else this proof can be done.

Theorem: If $G$ acts on $X$ then $\psi: X\to X$ defined by $\psi_g(x)=g\cdot x$ is a bijection for all $g\in G$.

Proof: If $x=y$ then $\psi_g(x)=g\cdot x = g\cdot y=\psi_g(y)$, thus $\psi_g$ is injective. Since $|im(\psi_g)|=|X|$, $\psi_g$ is bijective.

Answer 1

If $\psi_g(x)=\psi_g(y)$ then $gx=gy$. Then, applying $\psi _{g^{-1}}$ and using the associativity of the action, we have

$$\psi _{g^{-1}}(gx)=\psi _{g^{-1}}(gy)\Rightarrow g^{-1}(gx)=g^{-1}(gy)\Rightarrow (g^{-1}g)x=(g^{-1}g)y\Rightarrow x=y$$

so $\psi_g$ is injective.

If $x\in X$, then, again, using associativity, we have $\psi _g(g^{-1}x)=x$ which shows that $\psi _g$ is surjective.

Answer 2

Of course if $x=y$, then $\psi_g(x)=\psi_g(y)$! This is true for any function. What you need to show for injectivity is the converse of this, that is, if $\psi_g(x)=\psi_g(y)$, then $x=y$.

The second part about comparing cardinalities is correct (for finite sets $X$), but needs to be justified.