A categorical method to, say, determine the cardinality of a group

I am trying to figure out how much one can figure out about an object using category theory. Ideally, any property that is well defined up to isomorphism should be computable using only category theory. Let us say that we are trying to figure out how many elements are in a group? For a set, we could "simply" count the morphisms to it from the terminal object. Obviously, this wouldn't work. Is there a "categorical" method to find the cardinality of a group? If one wants a rigorous definition of what I mean by "categorically", here is your compass and straightedge:

1. You have an abstract symbol for the object in question.
2. For any two abstract symbols for objects $G$ and $H$, you can get abstract symbols for $Hom(G,H)$
3. For any abstract symbol for an object $G$, you can get the abstract symbol for $id_G$.
4. For any two abstract symbols for morphisms $f$ and $g$ such that $f \circ g$ is defined, you can get the abstract symbol of $f \circ g$
5. In terms of the objects and morphisms for abstract symbols you already have, you can get the abstract symbol of any object that uniquely (up to isomorphism) satisfies a given universal property (if needed, I can clarify this.)

Using a finite number of steps, I am trying to find the cardinality of the group. Note for example, you can't take the forgetful functor from $Grp$ to $Set$ (since you can't compute a functor on an abstract symbol.)

I am thinking one method would be to count how many automorphisms there are, but am I not sure how to get the cardinality of the group from this.

If these set of "rules" are too restrictive, it would be interesting to see how they could be lightened to make it possible.

• $|G| = |Hom(\mathbb{Z},G)|$ Jan 16 '15 at 1:09
• In a different, but similar question over at MSE math.stackexchange.com/questions/656279/… Martin Brandenburg and I gave a long list of examples how one can go about characterizing group theoretical notions in terms of the category of groups (in that particular case: the category of finite groups which is seems to be a harder case than the category of all groups) Jan 16 '15 at 8:29
• What does cardinality means if you only consider it internally to $\bf Grp$? Is the cardinality of $\Bbb{Z/6Z}$ and $S_3$ the same there? What about $\bigoplus_{i\in\Bbb Z}\Bbb{Z/2Z}$ and $\Bbb Z$ do they also have the same cardinality? And what if we compare those to $\bigoplus_{n\geq3}S_n$, what then? Jan 16 '15 at 22:41
• $hom(\mathbb{Z}, G)$ is not a group, it is a set of morphism. I would consider counting the morphisms between two objects "categorical." Jan 16 '15 at 23:09
• @AsafKaragila What else is "categorial" if not "doing all the stuff one can do with sets in the special case of hom-sets" ? Counting elements is stuff you can do with sets. It's almost the only stuff you can do with sets... Jan 17 '15 at 0:09

Every object $c$ in every category satisfies a universal property; it's universal for maps into or out of $c$, or in other words it represents the functors $\text{Hom}(-, c)$ or $\text{Hom}(c, -)$. So rule 5 certainly needs to be clarified. Once we have the ability to name any object we want we can of course just name $\mathbb{Z}$ and hence name the underlying set functor $\text{Hom}(\mathbb{Z}, -)$.

In any case, I think your conditions are much too restrictive. I think a natural example of a categorical method of answering this question would be to describe some categorical property which is satisfied uniquely by $\mathbb{Z}$ (but not necessarily a universal property).

In fact here is such a property: it's a theorem that the free groups are precisely the cogroup objects in $\text{Grp}$. $\mathbb{Z}$ is the unique such group which is indecomposable with respect to coproduct.

• Here is a proposal for a criterion by which one might give a negative answer to a question of this form: I would say that cardinality is not a categorical property if there were some automorphism of $\text{Grp}$ sending the underlying set functor $\text{Grp} \to \text{Set}$ to a non-isomorphic functor. The above argument shows this isn't the case; in particular it implies that every automorphism of $\text{Grp}$ fixes $\mathbb{Z}$. Jan 16 '15 at 7:00
• Moreover, it's an exercise in Freyd's Abelian Categories (pp. 31-32) that any automorphism of $\text{Grp}$ is uniquely isomorphic to the identity. See this answer by Reid Barton in a thread linked to already by dhy (in a comment under the original post). Jan 16 '15 at 9:02
• Continuing: Freyd's answer to Qiaochu's suggestion is that $\mathbb{Z}$ is the unique (up to isomorphism) group which admits more than one homomorphism to any nontrivial group and such that its only idempotents are the trivial ones $0$ and $1$. Jan 16 '15 at 9:10
• @PyRulez: the two conditions are separate. The first one is "admits more than one homomorphism to any nontrivial group" and the second one is "the only idempotent endomorphisms are the trivial ones." Jan 17 '15 at 0:37
• PyRulez, suppose $G$ is such a group. Then there is a nonzero homomorphism $G \to \mathbb{Z}$; since the image is isomorphic to $\mathbb{Z}$, there is a surjective $p: G \to \mathbb{Z}$. This $p$ admits a section, i.e., a homomorphism $i: \mathbb{Z} \to G$ such that $p i = 1_\mathbb{Z}$. Then $i p: G \to G$ is an idempotent, and clearly not the zero idempotent, and therefore $i p = 1_G$; thus we have proved $p: G \to \mathbb{Z}$ is an isomorphism. Jan 17 '15 at 1:19

A bunch of bits and pieces from a bunch of people:

Let us say we are trying to find the cardinality of the Group $G$. First, we select the group, $\bullet$ (unique up to isomorphism) such that for any other group, there is a unique arrow in and out of $\bullet$. Now, we select a group $\mathbb{Z}$, unique up to isomorphism, such that it only has two idempotent homomorhpisms, and it admits at least two morphisms to any other group besides $\bullet$ (thanks Todd Trimble). Now we find how many morphisms there are from $\mathbb{Z}$ to $G$ (special thanks to Steven Gubkin). This is $|G|$.

Proof

The group $\bullet$ is the trivial group. Now, the group $\mathbb{Z}$ of integers satisfies the properties above via the proof here. Now, for any other group $H$ which satisfies the properties, we know that there is a nontrivial morphism $f : H \rightarrow \mathbb{Z}$. Now the image of $f$ will be a group of integers, and so must be multiples of a given integer $n$ (which won't be zero since $f$ is nontrivial.) We can take a map from this to all the integers, so that we can turn $f$ into a surjection $\bar f$. Now, we make a morphism from $\mathbb{Z}$ to $H$, $i$, such that $i(1) = x$ for some $\bar f(x) = 1$.

Since $\bar f(i(n)) =\bar f(x+x+x+\dotsb)=\bar f(x)+ \bar f(x)+ \bar f(x)+\dotsb=1+1+1+\dotsb=n$, $\bar f \circ i = id_\mathbb{Z}$. This means that $i \circ \bar f$ is idempotent, and since $i(\bar f(x)) = x$, it is not the zero morphism. Since $H$ has only two idempotents (the zero morphism and $\operatorname{id}_H$), and $i \circ \bar f \neq 0$, $i \circ \bar f = id_H$. (Thanks Slade.) Therefore they are inverses. Therefore we can select $\mathbb{Z}$ up to isomorphism.

For each element of $G$, $x$ we make a morphism from $\mathbb{Z}$ to $G$, $h$, such that $h(n) = n * x$. Also, for any morphism $h: \mathbb{Z} \rightarrow G$, $h(n) = n * h(1)$, where $h(1)$ is an element of $G$. Therefore, the morphisms between the integers and $G$ are in one to one correspondence with the elements of $G$. Therefore $|\operatorname{Hom}(\mathbb{Z}, G)| = |G|$.

$\square$

• What do you mean, $\mathbb{Z}$ has "two idempotents"? As an additive group, $\mathbb{Z}$ has a unique idempotent, same as any other group: because $x^2=x$ implies $x=e$ (the identity). And multiplicatively, it is not a group. Jan 18 '15 at 19:54
• @ArturoMagidin I meant idempotent homomorphism, namely a homomorphism, say $f$, such that $f(f(x)) = f(x)$ for all $x$. I have clarified it now. (Category theorists think by default in homomorphisms instead of elements. Sorry for the confusion.) Jan 19 '15 at 1:21
• In the very first comment after your Question, Steven Gubkin said all there was to say (one could add the categorical characterization of $\ \mathbb Z,\$ but that's trivial). Jan 19 '15 at 2:00
• @WłodzimierzHolsztyński, to be fair, PyRulez did repeatedly ask StevenGubkin to turn his comment into an answer (starting with mathoverflow.net/questions/194047/…). Jan 19 '15 at 2:01
• @WłodzimierzHolsztyński Hm, what's the "trivial" categorical characterization of $\mathbb{Z}$ you had in mind? (Categorical characterizations of $\mathbb{Z}$ have been offered in other comments, but it's a slight stretch I think to call them "trivial". Remember that we're restricting ourselves to the first-order language of categories, interpreted in the category of groups.) Jan 28 '15 at 1:55

In the category of all groups, the epimorphisms are precisely the surjective mappings according to this encyclopedia of mathematics structures. Therefore, since the epimorphisms are surjective, one can recover the lattice of all normal subgroups of a group simply by taking all the epimorphisms where we regard two epimorphisms $f:A\rightarrow B_{1},g:A\rightarrow B_{2}$ as being equivalent if there is an isomorphism $i:B_{1}\rightarrow B_{2}$ with $if=g$ and we order the epimorphisms with domain $A$ where if $f:A\rightarrow B_{1},g:A\rightarrow B_{2}$, then $f\leq g$ if there is some epimorphism $j:B_{1}\rightarrow B_{2}$ with $jf=g$.

Similarly, the monomorphisms in the category of all groups are precisely the injective mappings, so one can recover the lattice of all subgroups of a group from the category of groups. If $G$ is a group, then the compact elements in the lattice of subgroups are precisely the compact elements in the lattice of subgroups of $G$. It is easy to see that a group is infinite if and only if it has infinitely many finitely generated subgroups: if $G$ is infinite and $G$ contains an element of infinite order $a$, then $\{\langle a^{n}\rangle|n\in\mathbb{N}\}$ are infinitely many finitely generated subgroups. If $G$ has no element of infinite order, then set $a\simeq b$ if $\langle a\rangle=\langle b\rangle$; then each equivalence class in $G$ has finitely many elements, so $G$ has infinitely many equivalence classes and hence infinitely many finitely generated subgroups.

I now claim that if $G$ is infinite, then there are $|G|$ many finitely generated subgroups of $G$. Clearly, there are $|G|$ many finite subsets of $G$, so there can be at most $|G|$ many finitely many subgroups of $G$. If we let $\simeq$ be the same equivalence relation as before where $a\simeq b$ iff $\langle a\rangle=\langle a\rangle$, then each equivalence class can have at most finitely many elements, so there are $|G|$ equivalence classes in $|G|$. Therefore $G$ has at least $|G|$ finitely generated subgroups. We therefore conclude that for infinite groups, the cardinality $|G|$ is equal to the number of compact elements in the lattice of all subgroups of $G$ which can be described in terms of categories.

• And then what do you do? Jan 16 '15 at 2:15
• The epimorphisms with domain $A$ form a preordered class $\mathcal{J}$. One can turn $\mathcal{J}$ into a partial order and in fact a complete lattice simply by letting two elements in $\mathcal{J}$ be equivalent if there is an isomorphism between the ranges preserving the original map. Jan 16 '15 at 2:18
• And then how do you get the cardinality? Jan 16 '15 at 2:21
• I don't see how the epimorphisms allow you to recover the lattice of all subgroups. If $G$ is a simple group, then the only epimorphisms with domain $G$ are those into the trivial group and those into $G$ itself. In the former case, there is but one epimorphism; in the latter case, you get $\mathrm{Aut}(G)$, and all maps are equivalent since given two automorphisms $f,g\colon G\to G$, we can let $i=gf^{-1}$ to get $if=g$, hence $f\leq g$ holds for any pair. So my lattice contains two elements: how did I recover the lattice of subgroups of $G$ from this? Jan 16 '15 at 4:23
• I was wondering... In any case: there is a nice, simple proof that all group epimorphisms are surjective due to Linderholm; you can see the proof here Jan 16 '15 at 4:54