Homology of a certain group Let $G$ be the group generated by $a_i,b_j$, $i,j=1,2,3$ and the following relations:
$$b_ib_j=b_jb_i, a_ib_j=b_ja_i, a_1a_2=b_3a_2a_1, a_2a_3=v_1a_3a_2, a_3a_1=v_2a_1a_3$$
I want to comute the homology of $H_nG=H_n(G,\mathbb{Z})$ where $\mathbb{Z}$ is considered as a trivial $G$-module.
I have made some attempts of applying spectral sequences (Lyndon-Hochschild-Serre). Here I chose $N=\langle b_1,b_2,b_3\rangle$ as the normal (central) subgroup of $G$. However I could not manage to compute the differentials. Does anyone know how to do this (either by spectral sequences or any other way)?
 A: I'll try to give an idea on how the computation can be done using the LHS spectral sequence (SS). I start computing cohomology, for cohomology admits a cup product in the SS that is quite useful. Let be $C := Z(G) = \langle b_1,b_2,b_3 \rangle \cong \mathbb{Z}^3$ and $Q := G/C \cong \mathbb{Z}^3$. 
Note that $H^\ast(C) = \Lambda_{\mathbb{Z}}(x_1,x_2,x_3)$ and $H^\ast(Q) = \Lambda_{\mathbb{Z}}(y_1,y_2,y_3)$. 
Since the extension 
$$ 1 \to C \to G \to Q \to 1\hspace{20pt}(\ast)$$
is central, by universal coefficients, 
$$E_2 = H^\ast(Q;H^\ast(C)) = H^\ast(Q) \otimes H^\ast(C) = \Lambda_{\mathbb{Z}}(x_1,...,y_3)$$
with $\deg x_i = (0,1)$, $\deg y_j = (1,0)$. Note that $d_2$ is a derivation, i.e. 
$$d_2(a \cdot b) = d_2(a)\cdot b + (-1)^{i+j}a \cdot d_2(b) \hspace{20pt}(\ast\ast)$$
for $a \in E_2^{i,j}, b \in E_2^{k,l}$. Since $d_2(y_j) = 0$ (for positional reasons), $d_2$ is completely determined by its values on the $x_i's$. 
Now there's an important formula for $d_2^{0,1}$. It depends on the extension class from $(\ast)$. A good reference is [Neukirch, Schmidt, Wingberg: Cohomology of Number Fields, Theorem 2.1.7]. Using the formula, I found
$$d_2(x_1) = y_1y_2,\hspace{2pt} d_2(x_2) = y_1y_3,\hspace{2pt} d_2(x_3) = y_2y_3.$$
Using $(\ast\ast)$ one can compute the $E_3$-term. In my (hasty) computation, I obtained 
$$E_\infty = E_3 = \begin{array}{cccc}
0 & 0 & \mathbb{Z}^3 & \mathbb{Z} \newline 
0 & \mathbb{Z}^6 & \mathbb{Z}^8 & 0 \newline 
0 & \mathbb{Z}^8 & \mathbb{Z}^6 & 0 \newline 
\mathbb{Z} & \mathbb{Z}^3 & 0 & 0
\end{array}$$
For positional reasons, $E_3 = E_\infty$ and since all groups are free abelian, it follows 
$$H^i(G) = 
\begin{cases}
\mathbb{Z} & i=0,6 \newline 
\mathbb{Z}^3 & i=1,5 \newline 
\mathbb{Z}^8 & i = 2,6 \newline 
\mathbb{Z}^{12} & i=3
\end{cases}$$
Finally, by duality, $H_i(G) = H^i(G)$. 
Perhaps someone can compare the result with the paper referenced by Mark Sapir. Thanks. 
A: Trivially $H_0G=\mathbb{Z}$.  And it's well known that $H_1G=G/[G,G]$.  And we have Hopf's formula $H_2G=R∩[F,F]/[F,R]$ for presentation $G=F/R$. So "low" dimensions are taken care of.  Computer programs (GAP,HAP) can compute group homology for a ton of finite groups and certain infinite groups, but I haven't used it in 3 years and am not sure if it works in your case.
A: If $v_i=b_i$, your group is the free nilpotent group of class 2 and rank 3. Its homology is known: http://iopscience.iop.org/1064-5616/189/4/A03. 
