Cayley’s Theorem

In the study of group actions at the elementary, there are two actions in particular that are very important. The first is the action of $G$ on itself by left-multiplication. The second is the conjugation action. On this page, we will be concerned with the former.

Suppose $G$ is a group acting on itself by left-multiplication. Something that we get literally for free is that the action is transitive. Why? Well, because we can just pick out $e$ as a representor for some orbit and then just act on $e$ by $g$, giving literally any element we want in the group. So $G$ has a single orbit.

We are particularly interested in when $G$ is a finite group of order $n$. In this case, we can label the elements of $G$ as $g_i$ for $i \in [n]$. Then the permutation representation in this case is the homomorphism $G \to S_n$ (abuse of notation) given by $g \mapsto \sigma_g$. And then $\sigma_g:[n] \to [n]$ is given by

That is, the representation is induced by the permutation of the indices. In addition, notice that the action is faithful since $gg_i = g_i$ implies $g$ is the identity by cancellation. Thus, $G$ is isomorphic to some subgroup of $S_n$. This gives a proof of the following theorem for the finite case.

Theorem (Cayley’s theorem). Every group is isomorphic to a subgroup of a symmetric group.

For the infinite case, we can easily modify the above argument so it is agnostic of $G$ having finite order.

Remark. In a certain sense, notice that Cayley’s theorem says that if we fully understand the symmetric group, then we more-or-less understand everything about abstract groups — this gives a hint at how difficult it is to fully understand something that is as seemingly simple as the symmetric groups, which are completely concrete.