Finitely generated subgroups of direct limits of groups 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.)
 A: 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.
A: 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 $\langle G|\{r_i|i\leq k\}\rangle$.
A: 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. 
