Let $G$ be the semidirect product of $\mathbb{Z}^2$ with $\mathbb{Z}/6$ where $\mathbb{Z}/6$ acts by the order 6 element of $SL\_2(\mathbb{Z})$. We can think of this group as the group of order preserving isometries of the tesselation of $\mathbb{R^2}$ with regular triangles.


*Does this group acts properly, isometrically and cocompactly on a median space??*


Let for two points in a metric space $[x,y]=\{z|d(x,z)+d(z,y)=d(x,y)\}$. If $X$ is a geodesic metric space than this is just the set of all points lying on some geodesic from $x$ to $y$. $X$ is called a median space if for every triple of points $x,y,z$ we have that $[x,y]\cap[x,z]\cap[y,z]$ consists of exactly one point - the median of $x,y,z$. Examples for median spaces are trees and $\mathbb{R}^n$ with the $l^1$- metric. 


The motivation is that the one skeleton of a CAT(0) cube complex is a median graph. If a group acts geometrically on this CAT(0)-cube complex it also acts that way on that graph.
For example this group acts properly and isometrically on $\mathbb{R}^3$. This gives a proper and isometric action on a median space, but this action is not cocompact.
So I was wondering whether there is a better action. The problem seems to be that the automorphism of $\mathbb{Z}^2$ does not extend to a cube-complex automorphism of $\mathbb{R}^2$, but I could not make this precise.



Does this group act geometrically on a Median space?
It is known that the group you mention does not act geometrically on a CAT(0) cube complex. See for instance [this answer](https://mathoverflow.net/questions/181863/cat0-groups-that-does-not-act-on-cat0-cubical-complex/295487#295487) for a possible argument, based on Lemma 16.12 in Wise's monograph *The structure of groups with a quasiconvex hierarchy*. I think you can adapt the proof of this lemma in order to prove the following statement:


**Proposition:** *Let $G$ be a group acting properly and cocompactly on a median metric space $(M,d)$ of finite rank. Assume that $G$ contains a finite-index subgroup $H \simeq \mathbb{Z}^n$, $n\geq 2$. Then $G$ acts properly and cocompactly on $(\mathbb{R}^n, \ell^1)$.*


**Sketch of proof.** As shown by Bowditch in *Some properties of median metric spaces*, there exists a CAT(0) metric $\sigma$ on $M$, and, if I understand the construction correctly, this metric satisfies the following properties: (1) any isometry of $(M,d)$ induces an isometry of $(M,\sigma)$; (2) the metrics $d$ and $\sigma$ are biLipschitz equivalent; (3) halfspaces of $M$ are $\sigma$-convex.


As a consequence, the flat torus theorem can be applied, and we find a $G$-invariant and $\sigma$-convex subspace $\Sigma \subset M$ which is $\sigma$-isometric to $\mathbb{R}^n$. 


Now, we consider the structure of measured wallspace of $\Sigma$ induced by the walls of $M$. By $\sigma$-convexity of $\Sigma$ and of the walls, we must have $m$ families of parallel hyperplanes $\mathbb{R}^{n-1}$ in $\Sigma$. (Here, we ignore the collections of hyperplanes which lie in the neighborhood of a single hyperplane. As a consequence, $m \leq n$ since otherwise it would be possible to embed coarsely $\mathbb{R}^{n+1}$ into $\Sigma \simeq \mathbb{R}^n$.) 


Let $F$ denote the median space associated to the previous wallspace. Then $F$ decomposes as the $\ell^1$-product of $m$ (discrete or continuous) unbounded lines. Up to replacing discrete lines with continuous lines, we may suppose that $F$ is $(\mathbb{R}^m,\ell^1)$.


Notice that, because $G$ acts properly on $\Sigma$, necessarily it also acts properly on $F$. So we must have $m \geq n$. But we already know that $m \leq n$, so $m=n$.


So far, we have proved that $G$ acts properly on $(\mathbb{R}^n,\ell^1)$. As $G$ is virtually $\mathbb{Z}^n$, we conclude that $G$ acts geometrically on $(\mathbb{R}^n,\ell^1)$.
$\square$


In your specific example, the question is now: does $\mathbb{Z}^2 \rtimes \mathbb{Z}\_6$ act geometrically on $(\mathbb{R}^2,\ell^1)$? The same argument as the one followed [here](https://mathoverflow.net/questions/181863/cat0-groups-that-does-not-act-on-cat0-cubical-complex/295487#295487) works. A presentation of the group is
$$T=\langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(bc)^3=(ac)^3=1 \rangle.$$
Notice that $\mathrm{Isom}(\mathbb{R}^2, \ell^1)= (\mathbb{R}\rtimes \mathbb{Z}\_2)^2 \rtimes \mathbb{Z}\_2$ does not contain elements of order three, so, for any action $T \curvearrowright (\mathbb{R}^2, \ell^1)$ by isometries, the elements $ab$, $bc$ and $ac$ must be trivial. Consequently, such an action must factorise through the quotient $T \twoheadrightarrow \mathbb{Z}\_2$ sending all the generators to $1$. A fortiori, the action cannot be geometric (and even proper).


**Remark:** The argument above does not completely answer the question as the median metric space is supposed to have finite rank. I do not know what happens if the rank is infinite, but it seems reasonable to think that the same conclusion holds.


