Group action induces homomorphism to the symmetric group

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.