Just as we can curry functions, so we can "curry" homomorphisms and actions.

Given an action of group $G$ on set $X$ , we can consider what happens if we fix the first argument to $ρ$ . Writing $ρ (g)$ for the induced map $X \to X$ given by $x \mapsto ρ (g, x)$ , we can see that $ρ (g)$ is a bijection.

Indeed, we claim that $ρ (g^{- 1})$ is an inverse map to $ρ (g)$ . Consider $ρ (g^{- 1}) (ρ (g) (x))$ . This is precisely $ρ (g^{- 1}) (ρ (g, x))$ , which is precisely $ρ (g^{- 1}, ρ (g, x))$ . By the definition of an action, this is just $ρ (g^{- 1} g, x) = ρ (e, x) = x$ , where $e$ is the group's identity.

We omit the proof that $ρ (g) (ρ (g^{- 1}) (x)) = x$ , because it is nearly identical.

That is, we have proved that $ρ (g)$ is in $S y m (X)$ , where $S y m$ is the symmetric group; equivalently, we can view $ρ$ as mapping elements of $G$ into $S y m (X)$ , as well as our original definition of mapping elements of $G \times X$ into $X$ .

$ρ$ is a homomorphism in this new sense

It turns out that $ρ : G \to S y m (X)$ is a homomorphism. It suffices to show that $ρ (g h) = ρ (g) ρ (h)$ , where recall that the operation in $S y m (X)$ is composition of permutations.

But this is true: $ρ (g h) (x) = ρ (g h, x)$ by definition of $ρ (g h)$ ; that is $ρ (g, ρ (h, x))$ because $ρ$ is a group action; that is $ρ (g) (ρ (h, x))$ by definition of $ρ (g)$ ; and that is $ρ (g) (ρ (h) (x))$ by definition of $ρ (h)$ as required.