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Let $G$ be a direct limit of groups $G_n$ for $n\in \mathbb{N}$ (or perhaps even for $n$ in some other directed set, but in my case I only need $\mathbb{N}$).

Is it true that every finitely generated subgroup of $G$ is isomorphic to a subgroup of some $G_n$?

[EDIT: I'm adding the category-theory tag in case it helps. I think category-theorists don't like the term 'direct limit' these days, so please read 'colimit' instead.)

Tara B
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  • I think this might be easy, but I'm really tired today and I really want to know the answer, but couldn't find it on the internet or find a suitable book in the library. – Tara B Apr 15 '13 at 16:17
  • I don't have a counterexample, but you might need finitely presented, not just finitely generated. – Jim Apr 15 '13 at 16:19

3 Answers3

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Let $(G_n)_{n \in N}$ be a directed system of groups and $(i_n : G_n \to G)_n$ their colimit. Let $H \leq G$ be a finitely presented subgroup. Then it is well-known that $H$ is finitely presentable, i.e. $\hom(H,-)$ preserves directed colimits (see Corollary 3.13 in LPAC). In particular $H \hookrightarrow G$ factors as $H \xrightarrow{j} G_n \xrightarrow{i_n} G$ for some $n$. We must have that $j$ is a monomorphism. Hence, $H$ is isomorphic to a subgroup of $G_n$.

For the sake of completeness and because it is so easy, here is a direct proof of this factorization: Say that $H$ is generated by $h_1,\dotsc,h_n \in H$, and that the only necessary defining relations are $R_1(h_1,\dotsc,h_n)=\dotsc=R_m(h_1,\dotsc,h_n)=1$, where $R_i$ are words on $n$ letters. Each $h_i$ lies in the image of some $G_m$. Since the system is directed, we can choose some large $m$ which works for all $h_i$. Say $h_i=i_m(g_i)$. Now make $n$ even larger: Each relation already holds in some $G_n$. Since there are only finitely many, they already hold in $G_n$ for some $n \geq m$. But this means precisely that $h_i \mapsto g_i$ extends to a well-defined homomorphism $H \to G_n$.

As you can see, it is crucial here that $G$ is finitely presented. If $G$ is only finitely generated, but not finitely presented, we can write $G$ as the directed colimit of finitely presented groups $G_n$ (by looking at the finite parts of a presentation of $G$), and we can take $H=G$ as a counterexample. There is no reason why $G$ should be isomorphic to a subgroup of $G_n$. In fact, when $G$ has no recursively enumerable set of relations, then Higman has shown (Subgroups of finitely presented groups, Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences, 1961) that $G$ does not embed into any finitely presented group.

Martin Brandenburg
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As Jim points out, the following argument is not quite complete. I'll leave it for now as a partial answer:

Let $g_1,\ldots,g_k$ be the finite generators of your finitely generated subgroup of $G$, then for every $i$, there is some $n_i$ such that $g_i\in G_n$ for every $n\geq n_i$. Define $n^\prime = \max_{i=1,\ldots,k} n_i$, then $g_i\in G_{n^\prime}$ for every $i\in\{1,\ldots,k\}$ and thus there is a homomorphism from a subgroup of $G_{n^\prime}$ to the subgroup of $G_n$.

The problem Jim points out seems to be the following:

Let $\langle G|R\rangle$ be a finitely generated group with countably many relations $R = \{r_i|i\in\mathbb{N}\}$. Consider the directed system of groups given by the natural homomorphisms $$\langle G|\{r_i|i\leq k\}\rangle\to\langle G|\{r_i|i\leq k+1\}\rangle$$ The direct limit of this system is $\langle G|R\rangle$, which is finitely generated but is not naturally a subgroup of any $\langle G|\{r_i|i\leq k\}\rangle$.

Abel
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  • Yes. Thanks. I hadn't even finished reading the definition of direct limit properly before I asked the question, to be honest! (I had been thinking about inverse limits yesterday, where I don't think the analogous result is true, and I hadn't quite realised direct limits are so much 'easier'). – Tara B Apr 15 '13 at 16:38
  • I'm tired enough that I'm not going to accept your answer yet even though I'm thoroughly convinced, just in case someone comes along and points out something wrong with it. I will in a few hours, though (i.e. next time I'm on the site). – Tara B Apr 15 '13 at 16:40
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    The $G_{n'}$ that you have chosen may contain all the generators of $G$, but it needn't contain all the relations between those generators. For example $\mathbb Z \to \mathbb Z/2 \to \mathbb Z/2 \to \cdots$ is a direct system with limit $\mathbb Z/2$. What, in your argument, stops me from taking $1 \in \mathbb Z$ as the preimage of the generator of $\mathbb Z/2$? Certainly $\mathbb Z/2$ is not a subgroup of $\mathbb Z$. – Jim Apr 15 '13 at 17:53
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    @Julien: Yes I changed my answer accordingly. It was too easy to be true :). Can you think of a concrete counterexample though? That would be great! – Abel Apr 15 '13 at 19:51
  • @Abel For some reason, the only direct limits I have really worked with had injective connections. So I have an awful intuition of the general case... – Julien Apr 15 '13 at 19:59
  • @Jim: Thanks for pointing that out. That's exactly what I was worried about, actually, but then Abel still somehow managed to convince me. – Tara B Apr 15 '13 at 20:37
  • Sorry for being so convincing :) – Abel Apr 15 '13 at 20:51
  • @Abel: Actually it's much more my fault for not getting enough sleep last night. – Tara B Apr 15 '13 at 21:05
  • Are you guaranteed to get an isomorphic subgroup here? It seems like you would get that it is a quotient of a subgroup of some $G_n$. –  Apr 15 '13 at 22:25
  • @SteveD Exactly, that's why the proof is incomplete. – Abel Apr 15 '13 at 22:26
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I think it's helpful to introduce some notation first: Let $(N,\le)$ be a directed set (i.e. for all $n,m\in N$ there is $k\in N$ with $k\ge n$ and $k\ge m$). Futher let $f_{nm}: G_n \to G_m\;(n\le m)$ be a directed system of groups. Then the colimit $G=\lim_n G_n$ exists and there are associated group homomorphisms $f_n:G_n \to G$ such that $f_m \circ f_{nm}=f_n\;(n\le m)$.

The colimit in the category of groups (and some other familiar categories like the category of modules over a ring) has the property that for each $g\in G$ there are $n\in N, g_n \in G_n$ with $g=f_n(g_n)$.

Now let $H\le G$ be finitely generated with generators $h_1,...,h_p$. There are $n_i \in N, g_{n_i}\in G_{n_i}$ with $f_{n_i}(g_{n_i})=h_i$. Choose $n\ge n_i\;(i=1,...,p)$. Hence $h_i=f_n(f_{{n_i},n}(g_{n_i}))\in f_n(G_n)$ and therefore

$$H \le f_n(G_n)$$

If all $f_{nm}$ are injective, $f_n$ is injective as well and $H$ is isomorphic to a subgroup of $G_n$ in this case.

Ralph
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