Is there a simple way of visualising the direct limit of the cyclic subgroups of a group?

Question

By way of background to this question, I am interested in the properties of direct limits. They are usually defined in terms that assume there is an underlying directed poset, but according to category theory, direct limits do in fact exist for general diagrams that are not directed.

I am interested in getting an intuitive feel for what direct limits which are not based on directed sets "look like". In other words, what do we lose, in not specifying the directed nature of the poset? Clearly we don't lose existence, since category theory tells us that the limits exist. So what is it?

Every group G is a direct limit (in the conventional sense) of its finitely generated subgroups, because any two such subgroups generate another f.g. subgroup, so there is a natural directed partial order on the set of f.g. subgroups. But the poset of cyclic subgroups is not directed, yet the direct limit must exist, and must contain G.

Can anyone give an example where the direct limits of the cyclic subgroups is not G itself, and in that case what is it?

For the broader question of why directedness is usually specified in defining direct limits (except in books on category theory), I suspect but cannot prove to myself that it is connected with the idea that the concept of a direct limit is in some way topological. Topologies introduce directed sets in a natural way, in that the basis of neighbourhoods of a point is directed downwards (the intersection of two neighbourhoods is another one contained on both). Can anyone put this vague intuition on a firmer basis (no pun intended)?

Answer 1

It's common to use the term "direct limit" for a directed colimit, so I'll use the more general term "colimit".

Take $G$ to be the Klein 4-group $C_2\times C_2$. The cyclic subgroups are the trivial subgroup and three cyclic subgroups of order 2, with the only (inclusion) maps being the identity maps and the inclusion of the trivial subgroup into the subgroups of order 2. In this case, the colimit (in the category of groups) of the cyclic subgroups is the free product $C_2\ast C_2\ast C_2$ of the three cyclic subgroups, which is an infinite group. It's not true that $G$ is contained in the colimit; rather, $G$ is a quotient of the colimit (the universal property of the colimit gives a natural map from the colimit to $G$ rather than an inclusion of $G$ to the colimit).

Answer 2

If $G : I \to \mathsf{Grp}$ is an arbitrary small diagram of groups with colimit diagram $(G_i \xrightarrow{\iota_i} \mathrm{colim}_i G_i)$, then the elements of $\mathrm{colim}_i G_i$ don't have to be of the form $\iota_i(g)$ for some $i \in I$ and $g \in G_i$. This is one of the main differences to the case that $I$ is a directed category. The most extreme example is when $I$ is discrete, because then the colimit is just the coproduct of groups. Any element of $\coprod_i G_i$ is a product $\iota_{i_1}(g_1) \cdot \dotsc \cdot \iota_{i_n}(g_n)$ with $g_k \in G_{i_k}$, where we may assume $g_k \neq 1$ and $i_k \neq i_{k+1}$. This gives a normal form for the elements of the coproduct. (This description of the elements has motivated many people to call this coproduct a free product, which is very unfortunate. Similarly, it is very unfortunate to call directed colimits "direct limits" ...) For general colimits, it is still correct that every element of $\mathrm{colim}_i G_i$ is such a product, but we get more relations and there is no general normal form for the elements.

Now for your question let $G$ be a group and $I$ be the partial order of its cyclic subgroups. We have a diagram $D : I \to \mathsf{Grp}$ which sends a cyclic subgroup to the corresponding group and every inclusion of cyclic subgroups to the corresponding monomorphism. Let us choose a colimit $(\iota_H : H \to G')_{H \in I}$. The universal property gives a homomorphism $f : G' \to G$ such that $f \circ \iota_H : H \to G$ is the inclusion for all $H \in I$. It follows easily that $f$ is surjective. But it doesn't have to be injective, even for abelian $G$:

Consider $G=\mathbb{Z}/2 \times \mathbb{Z}/2$. The partial order of cyclic subgroups has a minimal element, the trivial subgroup, and the three cyclic subgroups $\langle (1,0) \rangle$, $\langle (0,1) \rangle$, $\langle (1,1) \rangle$. There is no inclusion relation between them. Thus, $I$ is the discrete category on three objects with one initial object adjoined. It follows that $G'$ is the coproduct of the three mentioned groups, i.e. $\mathbb{Z}/2 \sqcup \mathbb{Z}/2 \sqcup \mathbb{Z}/2$ (in the category of groups), which is infinite and therefore cannot be isomorphic to $G$.

Edit: This is almost identical to the answer by Jeremy Rickard. We wrote our answers at the same time.