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 $\mathrm{Sym}(G)$.