Cayley's theorem

In group theory, Cayley's theorem, named in honour of Arthur Cayley, states that every group G is isomorphic to a subgroup of a symmetric group.^[1] More specifically, G is isomorphic to a subgroup of the symmetric group ${\displaystyle \operatorname {Sym} (G)}$ whose elements are the permutations of the underlying set of G. Explicitly,

• for each ${\displaystyle g\in G}$, the left-multiplication-by-g map ${\displaystyle \ell _{g}\colon G\to G}$ sending each element x to gx is a permutation of G, and
• the map ${\displaystyle G\to \operatorname {Sym} (G)}$ sending each element g to ${\displaystyle \ell _{g}}$ is an injective homomorphism, so it defines an isomorphism from G onto a subgroup of ${\displaystyle \operatorname {Sym} (G)}$.

The homomorphism ${\displaystyle G\to \operatorname {Sym} (G)}$ can also be understood as arising from the left translation action of G on the underlying set G.^[2]

When G is finite, ${\displaystyle \operatorname {Sym} (G)}$ is finite too. The proof of Cayley's theorem in this case shows that if G is a finite group of order n, then G is isomorphic to a subgroup of the standard symmetric group ${\displaystyle S_{n}}$. But G might also be isomorphic to a subgroup of a smaller symmetric group, ${\displaystyle S_{m}}$ for some ${\displaystyle m<n}$; for instance, the order 6 group ${\displaystyle G=S_{3}}$ is not only isomorphic to a subgroup of ${\displaystyle S_{6}}$, but also (trivially) isomorphic to a subgroup of ${\displaystyle S_{3}}$.^[3] The problem of finding the minimal-order symmetric group into which a given group G embeds is rather difficult.^[4][5]

Alperin and Bell note that "in general the fact that finite groups are imbedded in symmetric groups has not influenced the methods used to study finite groups".^[6]

When G is infinite, ${\displaystyle \operatorname {Sym} (G)}$ is infinite, but Cayley's theorem still applies.