# Group action induces homomorphism to the symmetric group

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.

Parents:

• Group action

“Groups, as men, will be known by their actions.”