# Cayley's Theorem on symmetric groups

Cayley’s The­o­rem states that ev­ery group $$G$$ ap­pears as a cer­tain sub­group of the sym­met­ric group $$\mathrm{Sym}(G)$$ on the un­der­ly­ing set of $$G$$.

# For­mal statement

Let $$G$$ be a group. Then $$G$$ is iso­mor­phic to a sub­group of $$\mathrm{Sym}(G)$$.

# Proof

Con­sider the left reg­u­lar ac­tion of $$G$$ on $$G$$: that is, the func­tion $$G \times G \to G$$ given by $$(g, h) \mapsto gh$$. This in­duces a ho­mo­mor­phism $$\Phi: G \to \mathrm{Sym}(G)$$ given by cur­ry­ing: $$g \mapsto (h \mapsto gh)$$.

Now the fol­low­ing are equiv­a­lent:

• $$g \in \mathrm{ker}(\Phi)$$ the ker­nel of $$\Phi$$

• $$(h \mapsto gh)$$ is the iden­tity map

• $$gh = h$$ for all $$h$$

• $$g$$ is the iden­tity of $$G$$

There­fore the ker­nel of the ho­mo­mor­phism is triv­ial, so it is in­jec­tive. It is there­fore bi­jec­tive onto its image, and hence an iso­mor­phism onto its image.

Since the image of a group un­der a ho­mo­mor­phism is a sub­group of the codomain of the ho­mo­mor­phism, we have shown that $$G$$ is iso­mor­phic to a sub­group of $$\mathrm{Sym}(G)$$.

Parents:

• Symmetric group

The sym­met­ric groups form the fun­da­men­tal link be­tween group the­ory and the no­tion of sym­me­try.

• I feel like sym­met­ric­group should be a req­ui­site for this page. How­ever, this page is linked in the body of sym­met­ric­group, so it seems a bit cir­cu­lar to link it as a req­ui­site. I think this situ­a­tion prob­a­bly comes up for most child pages; what’s good prac­tice in such cases?

• I think hav­ing it as a req­ui­site is best? I see the is­sue, but some peo­ple may ar­rive from other pages or search.