In extension to the question

Yoneda-Lemma as generalization of Cayley`s theorem?,

can someone point out to me where, in the categorical notation and analyzation of the Cayley's theorem, the symmetrc group of transformations of the group elements is?

Put another way: I see that the group structure is found under the set of all the transformation of the group elements and so the group is isomorphic to a subgroup of the symmetric group. But the symmetric group is a very big object (e.g. for $\mathbb{Z_3}$ with its 3 elements, $S_3$ has 3!=6 elements) and I don't find it in the Yoneda lemma.

Put another way: Cayley's theorem says something about a subset, basically that $G$ is isomorphic to some transformation on $G$, let's name it $\lambda(G)$, and the insight is $\lambda(G)\le S(G)$. As the Yoneda lemma only has an equal (or isomorphic) sign, I wonder where the $\le$ symbol is in the categorical language.

I can only assume that the lemma $F(A)=\text{nat}(\text{hom}(A,-),F)$ translates to $G=\lambda(G)$ if one plugs in the hom functor for $F$ and the resulution to my question would then be to see what the $S(G)$ is in terms of $\text{nat}(\text{hom}(A,-),F)$ and why. (I don't get much out of it as there seems to be only the one object $A$ I can play around. And I don't conceptualize natural transformations very good, I'm afraid.)


$S(G)$ is the set of all bijections in $\mathcal{Set}$ from $|G|$ to itself, where I denote by $|G|$ the underlying set of $G$. This is exactly the permutation group on $|G|$.

Now, the application of Yoneda that gets us Cayley is, as you say, taking $F=\hom_G(B,-),$ where $\hat{G}$ denotes the one-object category with $G$ as its arrows. Call its only object $X$. So, for every $A,B,$ we get $F(A)=\hom(B,A)=\textrm{nat} (\hom(A,-),\hom(B,-))$. But the only candidate for $A$ or $B$ is $X$, so all we really have is $\hom(X,X)=\textrm{nat} (\hom(X,-),\hom(X,-))$. Now by the construction of $\hat{G}, \hom(X,X)=G,$ so now we just want to see why the natural transformations from $\hom(X,-)$ to itself are contained in the bijections on $|G|$.

First, the objects of the image of $\hom(X,-)$ are just $\hom(X,X)=|G|.$ Let's interpret the images of morphisms in $\hat{G}$, which are the elements of $g,$ by the right action: $\hom(X,g): h \in |G| \mapsto hg$. Now a natural transformation $\alpha$ needs a component morphism at each object in the image of the functor, but since our functors have singleton images let's identify $\alpha$ with $\alpha_{|G|}$. The naturality $\alpha$ needs is given by this diagram: $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{ccc} |G|&\ra{\alpha}&|G|\\ \da{g}&&\da{g}\\ |G|&\ra{\alpha}&|G|\\ \end{array} $$ That is, we need $\alpha(hg)=(\alpha(h))g$ for each $h,g$. We can obviously accomplish this for a set of $\alpha$ isomorphic to $G$ by letting each $k$ in $G$ act on the left. Note that any natural $\alpha$ will have to be a bijection on $|G|,$ in short because the right action of $G$ on itself is transitive. So we see that the admissible bijections are some subset of $S(|G|)$; by Yoneda, they're exactly $G$, so that $G \leq S(|G|)$.

  • $\begingroup$ Okay I see now how the object-image of $\text{hom}(A,-)$ is the group ($\text{hom}(A,A)=G$) and how the morphism image is the set of left actions on the group $\text{hom}(A,f)=\lambda_f$, which is isomorphic to the $G$ too. Also I see that, by the Yoneda lemma, every element of $\text{nat}$ has an inverse. What I still not get is why the natural transformations relating $\text{hom}(A,-)$ with itself is an endomorphism of $G$. While the image of $\text{hom}(A,-)$ for the single object and every single morphism is $G$, the thing $\text{hom}(A,-)$ itself is a functor with an open argument slot. $\endgroup$ – Nikolaj-K Aug 13 '12 at 15:14
  • $\begingroup$ You should identify $\hom(A,-)$ with its image in $\mathcal{Set}$, that is, with $\hom(A,A)$ and morphisms sent over from $\hat{G}$. This is the right viewpoint because a natural transformation is defined entirely as a collection of morphisms in the image of our functor. $\endgroup$ – Kevin Carlson Aug 13 '12 at 19:24

$S(G)$ comes from $\mathbf{Set}$.

Let $\mathbf{G}$ be the category with one object * such that $\hom(*,*) = G$.

If you want to think of Cayley's theorem from a category-theoretic point of view, it's a combination of several distinct observations.

  • There is an obvious action of $G$ on itself
  • A $G$-set is the same thing as a functor $\mathbf{G} \to \mathbf{Set}$
  • The image of any monoid homomorphism from a group to a monoid is contained in the the units of the monoid (the submonoid of invertible elements)
  • The unit group of $\mathop{\text{End}}(X)$ is $S(X)$

The interesting part of the analogy between the Yoneda lemma and Cayley's theorem is not the last bullet point.

If you really want a more direct analog of a functor $G \to S(|G|)$, then observe that every arrow $f:Y \to Z$ of your category induces a function of sets

$$ f_* : \hom(X, Y) \to \hom(X, Z)$$

that is natural in $X$, and that $(fg)_* = f_* g_*$. (and there is a corresponding contravariant version)

If you really, really want something even more direct, I'm sure you can set up some sort of object and look at its endomorphism category, and have a functor from $\mathbf{C}$ \to that. I doubt it will be enlightening.


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