# 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.”