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.”