# Where is the symmetric group hidden in the Yoneda lemma?

In extension to the question

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 $G \leq S(|G|)$.

• 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. – Nikolaj-K Aug 13 '12 at 15:14
• 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. – 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.