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 \(\mathrm{Sym}(G)\).