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.)



Finitely presented group containing every $\mathrm{GL}_n(\mathbb{Z})$
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 K-theory, 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 "non-commutative 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 "non-commutative 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:


1. It shouldn't be hard to extend this to the ring $\mathbf{Z}[1/m]$. However, passing to rationals would seem much less easy.
2. All this is, at the level of handling presentation, is indeed quite analogous to the finite presentability of wreath products.


