I am completely stuck in the following linear algebra problem.


Consider a finite group $H$ and $N\times N$-matrices $M\_{g,h}$ with entries in $\mathbb{Z}$ for all $g,h\in H$. Assume $\sum\_{h\in H} M\_{g,h}=D$ for all $g\in H$, where $D$ is some $N\times N$-matrix with $\det(D)\neq 0$. Furthermore $M\_{g,h}=M\_{1,g^{-1}h}$ for all $g,h\in H$.


The QUESTION is wether the determinant of the $N|H|\times N|H|$-block matrix $M:=(M\_{g,h})\_{g,h\in H}$ is not zero.


For further reduction one may assume that the entries of the matrices are in the set $-1,0,1$, that $M\_{g,h,i,j}\neq 0$, $1\leq i,j\leq N$, implies $M\_{g,h',i,j}=0$ for all $h'\neq h$ and not more than three $M\_{g,h}\neq 0$ for fixed $g$.


The answer is certainly positive if $H$ is abelian, but unfortunately I cannot assume commutativity. 


I would be happy if someone knew a place where this kind of problem was dealt with before or could point out a solution. Of course I would appreciate a counterexample, although less satisfactory.



Nonsingularity of certain block matrices
Even if $H$ is Abelian (see [MO question](https://mathoverflow.net/questions/44954)), the answer is **No**. Here is a counter-example where $H=\mathbb Z\_2$ and $N=2$. We have $M=\begin{pmatrix} A & B \\\\ B & A \end{pmatrix}$, where $D=A+B$. The assumption is $\det(A+B)\ne0$. One verifies easily that $\det M=0$ if and only if $\det (A-B)=0$. Let us choose
$$A=\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \end{pmatrix},\qquad B=\begin{pmatrix} 0 & 0 \\\\ 0 & -1 \end{pmatrix}.$$
We do have $\det(A+B)\ne0$, but $\det(A-B)=0$, and therefore $\det M=0$. Yet the extra assumptions about the entries are satisfied too.


