Cayley's Theorem states that every group $G$ appears as a certain subgroup of the symmetric group $Sym\left(G\right)$ on the underlying set of $G$.

Formal statement

Let $G$ be a group. Then $G$ is isomorphic to a subgroup of $Sym\left(G\right)$.

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 $\Phi :G\to Sym\left(G\right)$ given by currying: $g\mapsto (h\mapsto gh)$.

Now the following are equivalent:

• $g\in ker\left(\Phi \right)$ 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 $Sym\left(G\right)$.