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)$~$.