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

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

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

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

That is, we have proved that \(\rho(g)\) is in \(\mathrm{Sym}(X)\), where \(\mathrm{Sym}\) is the symmetric group; equivalently, we can view \(\rho\) as mapping elements of \(G\) into \(\mathrm{Sym}(X)\), as well as our original definition of mapping elements of \(G \times X\) into \(X\).

\(\rho\) is a homomorphism in this new sense

It turns out that \(\rho: G \to \mathrm{Sym}(X)\) is a homomorphism. It suffices to show that \(\rho(gh) = \rho(g) \rho(h)\), where recall that the operation in \(\mathrm{Sym}(X)\) is composition of permutations.

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