# Group action induces homomorphism to the symmetric group

https://arbital.com/p/group_action_induces_homomorphism

by Patrick Stevens Jun 14 2016 updated Jun 14 2016

We can view group actions as "bundles of homomorphisms" which behave in a certain way.

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.