Martin Orr's Blog

Groups as categories

Posted by Martin Orr on Saturday, 02 May 2009 at 17:12

This post explains how we can consider groups as categories, along with treating the G-sets and G-homomorphisms I considered in my last post on group actions as category-theoretic objects. This is preparation for talking about the Yoneda lemma. Before reading this post, you will need to know the definitions of categories, functors and natural transformations.

For any group G, we can form a category corresponding to G: take a single object, which we'll call *, and a morphism (which must be * \to *) for each element of G. Then define composition of morphisms to be the same as multiplication in the group. (This is just an abstract version of my original concept of "active group".) This category might often just be called G, but to reduce confusion I shall (in this post at least) call it C.

The group axioms of associativity and identity imply the corresponding axioms for a category. The axioms for a category do not require inverses for morphisms, so we shall make the following definition:

A "group category" is a category with one object in which every morphism has an inverse.

Now every group corresponds to a group category as above. Conversely, given any group category, we can make a group out of its morphisms.

We can give category-theoretic interpretations of other concepts too: for example a G-set (i.e. an action of G) is the same as a functor C \to \mathrm{Set}. To see why, observe that in order to specify a functor S : C \to \mathrm{Set} you need to specify a single object S* in Set and a morphism of sets Sg : S* \to S* for each element g of the group, with the correct composition property. Now S* is the set on which G acts and Sg is the map x \mapsto g.x for each g.

And how about a natural transformation S \to T where S, T are both functors C \to \mathrm{Set} (corresponding to G-sets)? Well for that you just need to specify a morphism of sets u : S* \to T* such that u((Sg)x) = (Tg)u(x) for all g in G, x in S* ; this is precisely a G-homomorphism as defined last time.

A special type of functor we will need to consider is a (covariant) Hom-functor. Let C be any category. For each object x in C, we get a functor \mathop{\mathrm{Hom}}(x, -) : C \to \mathrm{Set}, which takes an object y to the set Hom(x, y) of morphisms x \to y, and takes a morphism f : y \to z to the function \mathop{\mathrm{Hom}}(x, y) \to \mathop{\mathrm{Hom}}(x, z) "postcomposition with f". (Properly we need a condition on the category to ensure that Hom(x, y) is actually a set for each x, y.)

In the case where C is a group category, there is only one object so only one covariant Hom-functor, \mathop{\mathrm{Hom}}(*, -). This takes * to the set of morphisms * \to * i.e. to the underlying set of the group G, and takes g in G to the set function G \to G : h \mapsto gh.

Translating into the language of G-sets and actions, \mathop{\mathrm{Hom}}(*, -) corresponds to the regular action of G on itself. So not only is this an important action, but it corresponds to an important functor.

Tags categories, groups, maths

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