What is the reason for stating Cayley's theorem this way?

Question

In my notes, Cayley's theorem reads:

Any group $G$ is isomorphic to a subgroup of $\text{Sym}\, X$ for some $X$.

On the other hand, several sources (such as Wikipedia) give a slightly more precise statement:

Any group $G$ is isomorphic to a subgroup of $\text{Sym}\, G$.

Is there a reason why we don't specify what $X$ is in the first case? I guess the "point" of the theorem is that any group can be related to the symmetric groups, so in some sense it doesn't really matter what $X$ is, but I'd like to be sure there is not something deeper under all this.

Answer 1

All that's important is that G embeds in the symmetries of a set X.

The set can be taken to be the underlying set of G, but the fact that G is a group does not come into play when defining Sym(G). (It does of course come into play when you seek to find an embedding of G into Sym(G) ). G could be replaced with any other set of the same (or larger) cardinality.

The representation you get when doing this is just one particular representation of G as a group of symmetries. Why would one ignore the rest? :) it's analogous to making a ring R a right module over itself. This leads to the regular representation of R, but R has many other representations aside deform the regular one.

One can also view such a representation of G as a functor of the category G (with one element) into the category of sets.

Answer 2

I would call the first theorem a corollary of the second theorem; the point is that it tells you that the "abstract" (via axioms) and the "concrete" (via a collection of permutations of a set closed under inverses and composition) definitions of a group agree.

As far as versions of Cayley's theorem go, though, I prefer this one:

Every group $G$ is precisely the group of automorphisms of $G$, regarded as a right $G$-set when equipped with right multiplication.

This tells you not only that every group embeds into a group of symmetries but that every group is a group of symmetries, in the sense of being an automorphism group of a set equipped with extra structure. The reason I prefer to state Cayley's theorem this way is that this is the statement that generalizes most directly to the Yoneda lemma.