Does there exist a concrete example of a finitely presented group that contains an isomorphic copy of $\operatorname{GL}_n(\mathbb{Z})$ for every $n\in\mathbb{N}$? I think the Higman embedding theorem implies such a group must exist, but probably not in an especially constructive way, so I'm curious if there's a concrete example. I'm also especially interested in whether there exists a type $F_\infty$ example. (I'm also generally curious if there are any interesting implications of such a group existing.)

4$\begingroup$ In fact, Higman proved something really strong: there exist universal groups that are finitely presented, a group being universal if it contains an isomorphic copy of every recursively presented groups. $\endgroup$– AGenevoisOct 11, 2021 at 16:53

1$\begingroup$ A naive question. It is known that $V$ does not contain $\mathbb{Z} \ast \mathbb{Z}^2$, so clearly $V$ is far from satisfying your condition. But $\mathbb{Z} \ast \mathbb{Z}^2$ embeds in $2V$, which in turn does not contain $\mathbb{Z}\ast\mathbb{Z}^3$ (if I remember correctly). Etc. So, given an $n \geq 1$, does there exists an $m \geq 1$ such that $mV$ contains $\mathrm{GL}(n,\mathbb{Z})$? $\endgroup$– AGenevoisOct 11, 2021 at 17:03

1$\begingroup$ What do you denote by $mV$? $\endgroup$– YCorOct 11, 2021 at 17:48

1$\begingroup$ @MattZaremsky I was indeed writing an answer while you replied. $\endgroup$– YCorOct 11, 2021 at 19:15

1$\begingroup$ Just wanted to clarify a small thing from the comment about $\mathbb{Z}*\mathbb{Z}^3$  there was indeed a conjecture that this shouldn't embed into $2V$, but it turns out that, apparently, $2V$ contains every RAAG, see arxiv.org/abs/2103.06663. (This is very surprising, at least to me.) So I'd say that technically I don't know for sure that $2V$ doesn't already contain every $GL_n(\mathbb{Z})$, though I seriously doubt it does (for example, I would already be surprised if $2V$ contains the Heisenberg group). $\endgroup$– Matt ZaremskyOct 12, 2021 at 23:55
2 Answers
The idea suggested by Anthony Genevois works.
I'm using that for any set $S$, the finitary linear group $\mathbf{Z}$ has the presentation with generators $e_{st}$ for distinct $s,t\in S$, and relators $[e_{pq},e_{qr}]=e_{pr}$ for any distinct $p,q,r\in S$, and $[e_{pq},e_{st}]=1$ for any distinct $p,q,s,t\in S$. If $\mathrm{St}_S(\mathbf{Z})$ is the group defined by this presentation (this is called Steinberg group), there is a canonical homomorphism into the finitary group $\mathrm{GL}_S(\mathbf{Z})$, mapping $e_{st}$ to the matrix with $1$ on the diagonal and at position $(s,t)$, and $0$ elsewhere. (Almost) by definition, the kernel is $K_2(\mathbf{Z})$ and is central. A classical theorem, which can be found in Milnor's book on Ktheory, is that the latter is of order $2$; let $w$ be the nontrivial element of the kernel, viewed as word on the $e_{st}$.
Now assume that $S$ is a $G$set, $G$ finitely presented, and that $G$ has finitely many orbits on $S^4$, and that stabilizers for the $G$action on $S^2$ minus diagonal are finitely generated. (There are many such examples: action of Thompson's groups on dyadics, natural action of Houghton groups, etc).
So we have the semidirect product $\tilde{H}=\mathrm{St}_S(\mathbf{Z})\rtimes G$ (and its quotient $H=\mathrm{GL}_S(\mathbf{Z})\rtimes G$, killing the central subgroup of order $2$ $\langle w\rangle$).
Then under the given assumptions, $H$ is finitely presented. Indeed, for simplicity suppose that $G$ has a single orbit on $S^2$ minus diagonal; fix a point and let $L$ be the stabilizer. Start from a finite presentation of $G\ast\mathbf{Z}$, with $\mathbf{Z}=\langle u\rangle$. This free product can be viewed as a "noncommutative wreath product" $G\ltimes \langle u\rangle^{*G}$, the action permuting free factors. Then, modding out by the relators $[s,u]$ whenever $s$ ranges over a finite generating subset of $L$, we obtain the "noncommutative permutational product" $G\ltimes \langle u\rangle^{*G/L}$ (also equal to $G*\mathbf{Z}/[L,\mathbf{Z}]$). The relators $[e_{pq},e_{qr}]=e_{pr}$ can be written as $[gug^{1},huh^{1}]=kuk^{1}$ for various values of $g,h,k$ in $G$. Since $G$ has finitely many orbits on $G^3$, only finitely many of these are enough and the other ones follow by conjugating. The same remark applies to relators $[e_{pq},e_{st}]$, using that there are finitely many orbits on $S^4$.
This shows that $\tilde{H}$ and hence $H$ is finitely presented. And it contains copies of $\mathrm{SL}_n(\mathbf{Z})$ for all $n$ (and hence $\mathrm{GL}_n(\mathbf{Z})$, as the latter embeds into $\mathrm{SL}_{n+1}(\mathbf{Z})$.
Notes:
It shouldn't be hard to extend this to the ring $\mathbf{Z}[1/m]$. However, passing to rationals would seem much less easy.
All this is, at the level of handling presentation, is indeed quite analogous to the finite presentability of wreath products.

$\begingroup$ I was taking a look at Milnor's book and was going to write an answer. But you are faster! It seems reasonable to think that examples of type $F_n$ can be constructed for arbitrarily large $n$, but what about examples of type $F_\infty$? (EDIT: After a thought, finally it should also be doable without too much trouble if it already works for arbitrarily large $n$.) $\endgroup$ Oct 11, 2021 at 19:23

$\begingroup$ Aha! Very nice. I would think if the action has finitely many orbits in $S^n$ for every $n$, and the stabilizer of any finite subset of $S$ is type $F_\infty$ (so like, $F$ acting on dyadics) then the resulting group is type $F_\infty$, but I would also think that this would take quite a lot more work to prove. $\endgroup$ Oct 11, 2021 at 19:28

$\begingroup$ Indeed the point seem to be essentially that relators of the different $\mathrm{SL}_n(\mathbf{Z})$ have "the same form", with all generators playing "the same role". We might expect something similar with presentation of (outer) automorphism groups of free groups? Mapping class groups? assuming there is a uniformly bounded number of "types" of generators and relators... $\endgroup$– YCorOct 11, 2021 at 21:24

$\begingroup$ @YCor Indeed, and actually in my original question I almost included, "I'm also curious about the analogous question using $Aut(F_n)$," but decided that was too many subquestions! $\endgroup$ Oct 11, 2021 at 23:24
Here is a justification of why there is a general principle behind the construction. (This a formal description of what I had in mind when writting my comment.)
Let $S$ be a set and $R$ a subset of the formal union $\bigcup_{k \geq 1} S^k \backslash \{ (s_1, \ldots, s_k) \mid \exists i \neq j, s_i=s_j\}$. For every $r \in R \cap S^k$, fix a word $w_r(x_1, \ldots, x_k)$ of $k$ variables on a fixed alphabet $\{x_1,x_2, \ldots\}$. Now, define the group $G$ by the presentation $$G(S,R,w_\cdot):= \langle S \mid w_r(r)=1, \ r \in R \rangle.$$ Next, assume that a group $H$ acts on $S$ by preserving $R$ (i.e. $hr \in R$ for all $r \in R$ and $h \in H$) and $w_\cdot$ (i.e. $w_{hr}=w_r$ for all $h \in H$ and $r \in R$). An element of $H$ permutes the generators of $G$, inducing a permutation of the words of generators, and finally an automorphism of $G$. Let's consider the associated semidirect product $G \rtimes H$.
Fact: The group $G \rtimes H$ is finite presented if the following conditions are satisfied:
 $H$ is finitely presented;
 $H$ acts on $S^n \backslash \{ (s_1, \ldots, s_n) \mid \exists i \neq j, s_i=s_j\}$ with finitely many orbits for every $n \leq \max \{ k \mid R \cap S^k \neq \emptyset\}$;
 for every $s \in S$, the $H$stabiliser of $s$ is finitely generated.
So far, I have just rephrased in a more general framework what we already said. The point I want to emphasize here is that many "big" groups can be embedded in finitely presented groups from this point of view.
Example 1: If $H \curvearrowright S$, the permutational wreath product $\mathbb{Z}/n\mathbb{Z} \wr_S H$ is an example. Here, $R=S \cup S^2 \backslash \mathrm{thick \ diagonal}$ and $w_s=x_1^n$, $w_{(r,s)}=[x_1,x_2]$.
This shows that $\bigoplus_\mathbb{N} \mathbb{Z}/n\mathbb{Z}$ embeds in a finitely presented group.
Example 2: An example I like is $\mathrm{Sym}_\mathrm{fin}(X) \rtimes H$ from an action $H \curvearrowright X$ (where $\mathrm{Sym}_\mathrm{fin}$ denotes the group of finitely supported permutations). Here, $S = X^2 \backslash \mathrm{thick \ diagonal}$, the action $H \curvearrowright S$ is induced by $H \curvearrowright X$, $R=S \cup S^2 \backslash \mathrm{thick \ diagonal} \cup \{ ((a,b),(b,c),(a,c)) \}$, and $w_s=x_1^2$, $w_{(r,s,t)} = x_1x_2x_1x_3$, $w_{(r,s)}=[x_1,x_2]$ if $r \cap s = \emptyset$ and $(x_1x_2)^3$ otherwise.
So $\mathrm{Sym}_\mathrm{fin}(\mathbb{N})$ embeds in a finitely presented group. (This also follows from the construction of Houghton groups.)
Example 3: More generally, the construction applies very well for families defined by (labelled) graphs, such as Coxeter and Artin groups. For instance, one can show that $\mathcal{B}_\infty$, the group of finitely supported braids with infinitely many strands, embeds in a finitely presented group. (This also follows from the construction of braided versions of Thompson's groups.)
Probably this can be done for other generalisations of braid groups (e.g. virtual braid groups, loop braid groups, twin groups).
Example 4: As already described by Yves, Steinberg groups can be also used, answering the main question of this discussion.

$\begingroup$ I'm not sure what you have in mind as regards the finitary symmetric group. I guess you want a presentation over all transpositions. Namely, given two distinct transpositions $s,t$, you prescribe $st$ to be of order $2$ or $3$ according to whether their supports intersect (say $s,t$ are adjacent). But this is certainly not a presentation: the Coxeter group defined by this is huge and contains free subgroups. One should at least prescribe $sts=u$ for any pairwise adjacent $s,t,u$. Also, I don't see the presentations you have in mind in case of braid groups? The naive one doesn't seem enough. $\endgroup$– YCorOct 12, 2021 at 16:32

$\begingroup$ I think when you use all transpositions (instead of adjacent transpositions), then instead of braid relations you have to use all those $sts=u$ things like you said, and then it successfully presents the finitary symmetric group (not sure of a reference). So the spirit of Example 2 seems right, if not the exact relations. $\endgroup$ Oct 13, 2021 at 0:01

$\begingroup$ Right, I forgot many relations! I agree that adding the $sts=u$ suffices, and the same trick works for braid groups. However, I am less confident that similar constructions work for other generalised braid groups. $\endgroup$ Oct 13, 2021 at 8:16