In what follows "homology" will mean group homology, i.e. $H\_\*(BG^\delta;{\mathbf R})$ for the group $G$ with the discrete topology.


It is well-known how to compute the homology of abelian groups, for example for a torsion-free abelian group A one has $H\_n(A)=A\wedge\_{\mathbf Z}A\wedge\_{\mathbf Z}\ldots\wedge\_{\mathbf Z}A$, with n factors.


Can one compute the homology of solvable groups, for example of the group
$$\left\{\left(\begin{array}{cc}a&b\\
0&a^{-1}\end{array}\right)\colon a,b\in{\mathbf C}\right\}$$
or of the group
$$\left\{\left(\begin{array}{cc}a&b\\
0&a^{-1}\end{array}\right)\colon a,b\in{\mathbf k}\right\}$$
where $k\subset{\mathbf C}$ is some number field?


If $G$ is a Lie group, Milnor's Comm.Math.Helv.'83-paper gives an isomorphism $H\_\*(BG^\delta;{\mathbf Z}/p{\mathbf Z})=H\_\*(BG;{\mathbf Z}/p{\mathbf Z})$, but it's not clear to me what to do with this and what might be the analogue for the number field case. 



Homology of solvable Lie groups made discrete
The example that you mention is the semidirect product for the multiplicative group $k^\times$ acting nontrivially on the additive group $k^+$. Its homology coincides with the homology of $k^\times$. To see this you can a Hochschild-Lyndon-Serre spectral sequence. The point is that $H\_i(Bk^\times;H\_j(Bk^+))=0$ when $j>0$ and $i\ge 0$. For this it suffices if $k^\times $ has a subgroup $T$ such that $H\_i(BT;H\_j(Bk^+))=0$ for all $i\ge 0$. Choose $T$ to be infinite cyclic, generated by a rational number $c\notin\lbrace 0,1,-1\rbrace$. $c$ acts on the rational vector space $H\_j(Bk^+)$ by $c^{2j}$, so that the homology of $BT$ is $H\_0=coker(c^{2j}-1)=0$, $H\_1=ker(c^{2j}-1)=0$, $H\_i=0$ for $i>1$.


