10
$\begingroup$

There are already a number of questions here asking "what do people mean when they say a group is a groupoid with one object?". A natural question to ask is "are there any interesting group theoretic facts that one can recover from the category theoretic study of a groupoid with one object?"

I am seeking as new examples besides the claim "Yoneda Lemma $\Rightarrow$ Cayley's Theorem" which is nice, but seems to be the only one people ever talk about. I am aware that one could use the groupoid definition to construct the free group on a finite set of generators, but that feels more like a tautology.

$\endgroup$
6
  • $\begingroup$ Would you be interested in a nontrivial fact about the category of groups, that I imagine is useful in group theory but don't know an application of? $\endgroup$ Commented May 21 at 19:01
  • $\begingroup$ @DaniëlApol Sorry, maybe I was rather imprecise with the working. I was moreso asking about group theory consequences when viewing groups as groupoids with one objects, not necessarily group theoretic facts obtained from category theory. $\endgroup$
    – Shrugs
    Commented May 21 at 20:12
  • 5
    $\begingroup$ not group theory proper, but generalizing from groups to groupoids allows a nicer formulation of the Seifert–Van Kampen theorem. $\endgroup$ Commented May 21 at 20:14
  • 2
    $\begingroup$ Not sure if this is the kind of thing you have in mind, but e.g. viewing a group as a category with one object inspires the bar construction by taking its simplicial nerve. $\endgroup$
    – Thorgott
    Commented May 21 at 21:11
  • 1
    $\begingroup$ @Thorgott I think this is the kind of thing I had in mind! $\endgroup$
    – Shrugs
    Commented May 21 at 22:35

2 Answers 2

14
$\begingroup$

One statement you can prove by viewing groups as one-object groupoids is that, for any group $G$, the product functor $G\times-\colon\mathrm{Grp}\to\mathrm{Grp}$ preserves connected colimits, so in particular pushouts and coequalizers. You can likely also check this by hand, but the category-theoretic argument I gave here (and which I copied to below) just uses that the category $\mathrm{Grpd}$ of groupoids is cartesian closed (as opposed to $\mathrm{Grp}$). For the reader's convenience, I copied and slightly adapted the argument below.

The category $\mathrm{Grpd}$ of groupoids is cartesian closed, and the inclusion $\mathrm{Grp}\to\mathrm{Grpd}$ of groups as one-object groupoids preserves connected colimits and all limits (in particular products). Namely, $\mathrm{Grp}$ is embedded as full subcategory, so any limit or colimit of one-object groupoids that results in a one-object groupoid also computes that limit or colimit in groups. The functor $\mathrm{ob}\colon\mathrm{Grpd}\to\mathrm{Set}$ that takes the set of objects of a groupoid has both adjoints (exercise), so the set of objects of a limit or colimit of groupoids is computed as the limit or colimit of the underlying sets of objects of the groupoids involved. Any limit of singleton sets and any connected colimit of a diagram of singleton sets result in another singleton set, and this proves that $\mathrm{Grp}\to\mathrm{Grpd}$ preserves all limits and all connected colimits, and we can even compute those limits and colimits of groups inside the category of groupoids. Cartesian closedness of $\mathrm{Grpd}$ therefore implies that $G\times -\colon\mathrm{Grp}\to\mathrm{Grp}$ preserves all connected colimits.

$\endgroup$
2
  • $\begingroup$ Nice argument! +1 $\endgroup$
    – user43208
    Commented May 21 at 20:39
  • $\begingroup$ I am very satisfied with this particular example. Thanks for answering! $\endgroup$
    – Shrugs
    Commented May 21 at 22:12
7
$\begingroup$

$\newcommand{\B}{\mathbf{B}}$Another fascinating thing is Mackey's formula. In fact this groupoidal view is the only way I have to this day of remembering the details of the formula at all! As a notational distinction, I'll write the group as $G$ and the 1-object groupoid as $\B G$. A representation of a group is a functor $\B G \to \mathsf{Vect}$, so the functor category $\B G \to \mathsf{Vect}$ is the category of representations $\newcommand{\Rep}{\operatorname{Rep}}\Rep(G)$. (Let's restrict to finite groups and finite-dimensional vector spaces for simplicity.) Given a group homomorphism $H \to G$, we can compose it with a representation to get the restriction functor $\newcommand{\Res}{\operatorname{Res}}\Res^G_H : \Rep(G) \to \Rep(H)$. It has an adjoint called induction $\newcommand{\Ind}{\operatorname{Ind}}\Ind^G_H : \Rep(H) \to \Rep(G)$.

Suppose we have two subgroups $H \subseteq G \supseteq K$, Mackey's formula lets us compute $$\Res^G_H \Ind^G_K W = \bigoplus_{HgK \in H\backslash G/K} \Ind^H_{K_g} \Res^K_{K_g} W$$ for $W : \Rep(K)$. Here $K_g = H \cap g K g^{-1}$.

This is a well-known result in representation theory, but it is actually a Beck–Chevalley condition of a pullback square! The Beck–Chevalley condition is a phenomenon appearing in a wide range of mathematics including logic and geometry, where you have a pair of adjoint functors $f^* \dashv f_*$ associated with each morphism $f$ in a category: Given a pullback square $$\begin{matrix}A & \longrightarrow & B \\ \downarrow && \downarrow \\ C & \longrightarrow & D \end{matrix}$$ we can go from $B$ to $C$ in two ways: taking the upper left route or the lower right route (recall that adjoins go in opposite ways, so you take the appropriate one to compose), and the BC condition says they are isomorphic.

In logic, these two functors are $\exists_f \dashv f^*$ or $f^* \dashv \forall_f$, and the condition states that substitution commutes with quantification. In geometry the functors are pushforward and pullback. And in representation theory, the functors are restriction and induction. Consider the pullback $\B H \to \B G \leftarrow \B K$, whose objects are triples $(X, Y, e)$, where $X$ is an object of $\B H$, $Y$ of $\B K$, and $e$ an isomorphism of the corresponding images in $\B G$. (This is just pullback of sets categorified to categories/groupoids.) If you compute this pullback, you get a groupoid that is equivalent to the disjoint union of $\B K_g$, where $g$ ranges over representatives of the double coset $H\backslash G/K$. (It's a good group-theoretic exercise, try it!) This situation generalizes to Mackey functors, important in equivariant mathematics.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .