# Why is every abelian group the direct limit of its finitely generated subgroups?

I'm taking classes in homological algebra now, and the book (together with the lecturer) seem to assume more category theory than I already know.

A "fact" that is used freely in the book ("Homological algebra" by Weibel, by the way), is that every abelian group is (isomorphic to) the direct limit of its finitely generated subgroups.

However, I do not find this particularly obvious. How does one go about prove such things?

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Dear Fredrik, Based on your comment to Qiaochu's answer, it may help psychologically to point that the same is true if one replaces finitely generated subgroups by countably generated subgroups, or even by all subgroups. The point is that the statement you ask about (that a gp. is the direct limit of its f.g. subgps.) is stronger than these other statements, and (as Pete Clark points out in his comment following yours) allows one to reduce certain questions from the case of arbitrary abelian groups to the case of finitely generated abelian groups. –  Matt E Jan 31 '11 at 2:46

The fact that a group is the direct limit of its finitely generated subgroups is true for any group, not just abelian ones.

Remember what the direct limit of groups is. You start with a directed set $I$ (directed means that it is a partially ordered set, and if $i,j\in I$, then there exists $k\in I$ such that $i\leq k$ and $j\leq k$). For each $i\in I$, you have a group $H_i$, and if $i\leq k$, then you have a morphism $f_{ik}\colon H_i\to H_k$. The system $\{H_i, f_{ij}\}_{i\in I}$ must also satisfy:

1. $f_{ii} = \mathrm{id}_{H_i}$ for each $i\in I$; and
2. If $i\leq j\leq k$, then $f_{ik} = f_{jk}\circ f_{ij}$.

The direct limit is then constructed as follows: we let $X$ be the disjoint union of the sets $H_i\times\{i\}$, and we define the equivalence relation $\sim$ on $X$ as follows: given $(h,i)$ and $(h',j)$, we let $(h,i)\sim (h',j)$ if and only if there exists $k\in I$ such that $i\leq k$, $j\leq k$, and $f_{ik}(h) = f_{jk}(h')$.

We then define a group operation on $X/\sim$ as follows: given $[(h,i)]$ and $[(h',j)]$ in $X$, let $k\in I$ such that $i\leq k$ and $j\leq k$. We define $$[(h,i)][(h',j)] = [(f_{ik}(h)f_{jk}(h'),k)].$$ Note that this makes sense, since $f_{ik}(h)$ and $f_{jk}(h)$ are both in $H_k$. It is then an exercise to show that this gives you a group with the appropriate universal property.

Intuition. The idea is just that two elements in $X$ are "equal" if and only if they "eventually" map to the same thing. And we define the product by first mapping both elements sufficiently "ahead" that they both lie in the same group, and then multiply them. The direct limit is then a way of "gluing" all of $H_i$ in a compatible way.

The reason we expect the direct limit of the finitely generated subgroups to "be" the group (that is, isomorphic to the group) is that any particular operation we want to do with group elements always occurs in a finitely generated subgroup: if you are performing an operation inside the group, the operation uses only finitely many elements, so it all happens inside a finitely generated group. Thus, everything that "determines" what the group is is captured if you look at all finitely generated subgroups: to know how to multiply $x$ by $y$, you can multiply them in the finitely generated subgroup $\langle x,y\rangle$, after all. The direct limit is just a way of putting together all of these subgroups which we should be able to do since they are all "really" already glued together inside of $G$.

Sketch of proof. Now, let $G$ be a group. Let $I$ be the collection of all finitely generated subgroups of $G$, and order $I$ by inclusion of subgroups. For each element $i\in I$, let $H_i$ be the subgroup $i$ itself (remember that $i$ is a subgroup of $G$ by definition of $I$). If $i\leq j$, then $i\subseteq j$ as sets, so we let $f_{ij}\colon H_i \to H_j$ be the inclusion map. Note that if $i=j$, then $f_{ii} = \mathrm{id}_{H_i}$, as required, and if $i\leq j\leq k$, then $H_i\subseteq H_j\subseteq H_k$, and the inclusion $H_i\hookrightarrow H_k$ is the composition of the inclusion $H_i\hookrightarrow H_j$ and $H_J\hookrightarrow H_k$. So $\{H_i,f_{ij}\}_{i\in I}$ is a directed system of groups. Let $H$ be the direct limit, and let $f_i\colon H_i\to H$ be the structure maps into the direct limit. The structure maps are one-to-one, because $[(h,i)] = [(e,j)]$ if and only if there exists $k\geq i$ such that $f_{ik}(h) = e$, but all our $f_{ik}$ are one-to-one.

To see that the direct limit is (isomorphic to) a subgroup of $G$, note that you have embeddings $\varphi_i\colon H_i\hookrightarrow G$ from $H_i$ to $G$ for each $i$, and this embeddings commute with the structure maps $f_{ij}$. This means that by the universal property there is a homomorphism $\phi\colon H \to G$ such that $\varphi_i = \phi\circ f_i$ for all $i$. In particular, $\phi$ is an embedding, so $H$ is (isomorphic to) a subgroup of $G$. To see that the map is onto, let $g\in G$. Then letting $i=\langle g\rangle$, we have that $g$ is in the image of $\varphi_i$, hence in the image of $\phi$. Thus, $\phi$ is an isomorphism.

Added. The same argument holds for any collection $\mathcal{C}$ of subgroups of $G$ such that (i) every element of $G$ lies in at least one subgroup in $\mathcal{C}$; (ii) given any two subgroups $H$ and $K$ in $\mathcal{C}$, there is a subgroup $M$ in $\mathcal{C}$ that contains both $H$ and $K$. So it works for $\mathcal{C}$ being "all finitely generated subgroups"; "all subgroups"; for infinite cardinal $\kappa$, "all subgroups of cardinality less than or equal to $\kappa$"; and other classes.

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I tend to prefer doing this by using as indexing set the set of finitely generated subgroups itself: the apparent circularity is cute :) –  Mariano Suárez-Alvarez Jan 31 '11 at 1:16
@Mariano: Sure, this has the disadvantage that the same subgroup may occur several times. I'll redraft. –  Arturo Magidin Jan 31 '11 at 1:17
Tiny nitpick: your first statement appears to imply that colimits in Grp of abelian groups agree with colimits in Ab. This is false (for coproducts, for example): what is true is that Ab is reflective, so abelianization preserves colimits. –  Qiaochu Yuan Jan 31 '11 at 1:22
@Qiaochu: It's false that colimits in Grp of abelian groups agree with colimits in Ab: the coproduct of two abelian groups in Grp is not equal to the coproduct of two abelian groups in Ab (free product vs. direct sum). Here, the resulting group is abelian (since it is isomorphic to $G$), so there is no problem. –  Arturo Magidin Jan 31 '11 at 1:24
@Arturo: sorry, yes, I corrected my comment. –  Qiaochu Yuan Jan 31 '11 at 1:25

In nice algebraic categories like $\text{Ab}$, the colimit (direct limit) over a (edit: directed) diagram of inclusions behaves like a "union" operation ought to, so this result shouldn't come as a surprise. In any case, let's verify the universal property. Given an abelian group $G$, let $G_i, i \in I$ be an indexing of its finitely generated subgroups, which fit together into the obvious diagram of inclusions. We want to verify the following universal property: given an abelian group $H$ and a compatible system of morphisms $\phi_i : G_i \to H$, these morphisms factor through a unique morphism $\phi : G \to H$.

This is not actually a very deep fact. A compatible system of morphisms $\phi_i$ is precisely a list of where every element of $G$ goes (since every element of $G$ is contained in a finitely generated subgroup, namely the one it generates; this already proves uniqueness) together with a guarantee that this list respects all the relations of $G$ (since relations involve a finite number of elements, so can be verified in the subgroup they generate). In other words, it defines a unique morphism $G \to H$.

Again, in nice algebraic categories the same is true: for example, it's true of groups, rings, commutative rings, .... This has the following nice application: since a commutative ring is the colimit of its finitely-generated subrings, an affine scheme is the limit of Noetherian schemes (and this is still true relative to a Noetherian base). It also implies that the spectrum of a Boolean ring is a limit of finite sets, hence a profinite set (that is, a Stone space); see, for example, this blog post.

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Oh, such verifications are that straightforward. But wouldn't the same proof work if we let I be all subgroups of G? –  Fredrik Meyer Jan 31 '11 at 1:22
@Fredrik: yes, and the statement is still true (but not very meaningful: if you're taking a colimit over a diagram of inclusions with an object containing all the others (namely G), everything is already determined by that object). –  Qiaochu Yuan Jan 31 '11 at 1:28
Your first sentence is true of colimits of diagrams of inclusions over a directed poset. –  Mariano Suárez-Alvarez Jan 31 '11 at 1:39
@Fredrik: perhaps it will help to see how the fact you were asking about is actually used. The idea is that it one hopes in this way to reduce certain properties of arbitrary abelian groups (which are very complicated) to finitely generated abelian groups (which are completely understood). –  Pete L. Clark Jan 31 '11 at 1:43
@Mariano: thanks for the correction. –  Qiaochu Yuan Jan 31 '11 at 1:44