If one views a group as a one object category with the elements of the group as morphisms then a natural transformation between functors of such categories is an inner automorphism, i.e. if we have two group homomorphisms $f,g: A\to B$ then a natural transformation $\eta :f\to g$ is just an element $b\in B$ such that $f(a)\cdot b = b \cdot g(a)$ which can be rewritten as $f(a)=b \cdot g(a)\cdot b^{1}$. This isn't the only way to turn groups into categories. Another way is to take the elements of the group as objects and to have a morphism $h_a:a\to b$ if $h\cdot a=b$. If we view groups in this way then are the natural transformations again something nice like inner automorphisms?

6$\begingroup$ David  given the question you want to ask, I think it would be fair to tell us first what functors are between groups viewed in this second way. But when you've figured that out, you may find that your question has answered itself.... $\endgroup$ – Tom Leinster Feb 6 '10 at 4:54

4$\begingroup$ David, I don't think so. Given a group G, let's write C_G for the second resulting category that you mention ("Another way is..."). Write 1 for the trivial group. Then C_1 is the category with only one object and only the identity arrow. If G is any nontrivial group, then C_G has more than one object, so there is more than one functor from C_1 to C_G. On the other hand, there is only one homomorphism from 1 to G. $\endgroup$ – Tom Leinster Feb 6 '10 at 5:27

4$\begingroup$ Am I confused, or in the second method, do the morphisms have nothing to do with the objects? The way I'm reading it, between any two objects, there is precisely one morphism. $\endgroup$ – Steve D Feb 6 '10 at 5:49

2$\begingroup$ @Cory: Sorry, I should've been clearer. In the first construction, each group element is a morphism, and so the functors are group homomorphisms. In the second construction, while each morphism is labeled by group elements, there is nothing forcing functors to be homomorphisms, even though on these labels we get f(ba') = f(b)f(a)'. $\endgroup$ – Steve D Feb 6 '10 at 6:36

2$\begingroup$ @Cory: Ah, but in a preorder, the Homsets do carry information: either there is a morphism or there isn't. In C_G, there is precisely one morphism, so this is a very boring category indeed. $\endgroup$ – Theo JohnsonFreyd Feb 6 '10 at 7:38
The comments thread is getting a bit long, so here's an answer. The category $C(G)$ that David associates to a group $G$ (by his second recipe) has the elements of $G$ as its objects, and exactly one morphism between any given pair of objects. It's what category theorists call an indiscrete or codiscrete category, and graph theorists call a complete graph or clique. You can form the indiscrete category on any set: it doesn't need a group structure.
A functor from one indiscrete category to another is simply a function between their sets of underlying objects. In particular, given groups $G$ and $H$, a functor from $C(G)$ to $C(H)$ is simply a function from $G$ to $H$. That's any function (map of sets) whatsoever  it completely ignores the group structure.
Given indiscrete categories $C$ and $D$ and functors $P, Q: C \to D$, there is always exactly one natural transformation from $P$ to $Q$. In particular, given groups $G$ and $H$ and functors $P, Q: C(G) \to C(H)$, there is always exactly one natural transformation from $P$ to $Q$.

$\begingroup$ A related question was asked by Adam Libster on Feb 3. I gave some references there that might be of interest. $\endgroup$ – Tim Porter Feb 6 '10 at 7:25


1$\begingroup$ There is a strong link with ideas of nonAbelian cohomology for which see work of Larry Breen on the ArXiv (probably his notes for the IMA (BaezMay) meeting are the clearest). There is some great stuff by Aldrovandi and Noohi (check on their web pages involving butterflies! Yes that is right!) The earlier papers by Noohi are nice for that and if you want, raid my nLab home page, go to the personal page from that and download the Menagerie. I am writing a revised account of the automorphism 2group of a group to insert there, but I keep on being distracted by nice MO questions. $\endgroup$ – Tim Porter Feb 6 '10 at 8:03
I like the notation $\mathcal{B}G$ and $\mathcal{E}G$ for the two constructions of a category out of a group $G$ in David's question. $\mathcal{E}G$ is what Tom calls the codiscrete category $C(G)$.
Of course there is a third construction: it has $G$ as the objects, and only identity morphisms. Let's denote this category again by $G$.
The notation is nice because you can take the nerve of any category $\mathcal{C}$, and then geometrically realize. If we denote the resulting space by $\mathcal{C}$,
 $\mathcal{B}G$ is a classifying space for $G$.
 $\mathcal{E}G$ is a universal principal $G$bundle
 $G$ is just $G$.

$\begingroup$ I may not really understand the "geometric realization" functor so my question is irrelevant, but does this work even for topological groups? $\endgroup$ – Aaron MazelGee Feb 6 '10 at 20:21

$\begingroup$ @Aaron: For topological groups you probably want to take into account the topology, and Konrad's constructions don't do that. One can ask, though, if these constructions can be changed so as to take a topology on the group into account... $\endgroup$ – Mariano SuárezÁlvarez Feb 6 '10 at 20:37

$\begingroup$ Mariano is right, there is a geometrical realization functor that takes the topology on a category into account. With this modification, the answer to Aaron's comment is yes, at least for a fairly large class of topological groups. The reference is G. Segal "Classifying spaces and spectral sequences". $\endgroup$ – Konrad Waldorf Feb 6 '10 at 20:41