Cayley's Theorem states that every group appears as a certain subgroup of the symmetric group $S y m (G)$ on the underlying set of $G$ .

Formal statement

Let $G$ be a group. Then $G$ is isomorphic to a subgroup of $S y m (G)$ .

Proof

Consider the left regular action of $G$ on $G$ : that is, the function $G \times G \to G$ given by $(g, h) \mapsto g h$ . This induces a homomorphism $Φ : G \to S y m (G)$ given by currying: $g \mapsto (h \mapsto g h)$ .

Now the following are equivalent:

$g \in k e r (Φ)$ the kernel of $Φ$

$(h \mapsto g h)$ is the identity map

$g h = h$ 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 $S y m (G)$ .