How to prove Cayley's theorem using Category Theory

Question

I am working through the Steve Awodey's book "Category Theory" and there is the Cayley's theorem (Every group $G$ is isomorphic to a group of permutations) left for the reader to prove with this sketch of the proof (the page 13):

1. Define the group $\bar G$ of permutations as follows: $\forall g \in G$, there is a permutation $\bar g: G \to G$ defined $\forall h \in G$ by "acting on the left" $$\bar g(h) = g \cdot h$$$\checkmark$I can easily prove that $\bar G$ is indeed a group.
2. Define the homomorphisms: $i: G \to \bar G$ by $i(g) = \bar g$, and $j : \bar G \to G$ by $j(\bar g) = \bar g(1_G)$. $\checkmark$ Check
3. Show that $i \circ j = 1_\bar G$ and $j \circ i = 1_G$

I approach the third point like this: $$\forall g \in G, \space (j \circ i) \space g =\space j\space(i\space g) = j\space \bar g = \bar g\space1_G = g \circ 1_G = g$$So I get back to $g$, since there is only one identity morphism I can conclude that $j \circ i = 1_G$.

The inverse case ($i \circ j = 1_\bar G$) is proven symmetrically.

What seems to be off for me in this proof is that I deal with elements of the group, which I am not supposed to do. Is my intuition correct? Am I missing something? What is the canonical way to prove this theorem using the category theory?

The author defines a group as a monoid where every morphism is isomorphism, is there a way to use this fact in the proof?

Answer 1

Let me sum up the discussion in the comments.

This proof that is sketched isn't particularly categorical but it does relate to the Yoneda embedding $\mathcal{C} \to \mathsf{Set}^{\mathcal C^{\rm op}}$ where we map objects $A$ of $\mathcal{C}$ to their hom-functors $h^{A} : \mathcal{C}^{\rm op} \to \mathsf{Set}$. My guess is that the authors intend to explain the categorical connection later.

In the case were $\mathcal{C}$ is a group with a single object $A$, then $\mathcal{C} = \mathcal{C}^{\rm op}$ and $h^A : \mathcal{C} \to \mathsf{Set}$ takes $A$ to $\operatorname{Hom}(A,A) = G$ and group elements $g : A \to A$ to $\mathsf{Set}$-morphisms $\bar g(h) = g \circ h$. So this gives a map from $G = \operatorname{Hom}(A,A) \to \operatorname{Perm}(G)$.

The subgroup of $\operatorname{Perm}(G)$ which is the image of $\operatorname{Hom}(A,A)$ is the group of natural transformations of $h^A$. A natural transformation of $h^A$ looks like

$$\require{AMScd} \begin{CD} A @>g>> A\\ @V h^A V V @VV h^A V\\ \operatorname{Hom}(A,A) @>>\bar g> \operatorname{Hom}(A,A) \end{CD} $$

You can see this all on Wikipedia. Hopefully the above gives you some categorical context for Cayley's thoerem.

Answer 2

There nothing wrong with looking at elements of the group. If a group is seen as a one-object category, then what would be thought of as elements in traditional abstract algebra now become identified as the morphisms of the category. And one is certainly “allowed” to deal with elements of morphism sets.