Invertible matrix with group ring coefficient

Question

Before asking the question I do need some notations.

1. $G$ a (torsion-free) group, $\mathbb{Z}^{´}=\mathbb{Z}[\frac{1}{2}]$
2. $R:= \mathbb{Z}[G]$, $R^{´}=\mathbb{Z}^{´}[G]$ group rings.
3. $Mat_{n}(R)$ the ring of matrices of size $n\times n$, $GL_{n}(R)$ the group of invertible matrices.
4. $h: R\rightarrow \mathbb{Z}$ the augmentation ring homomorphism.

Is there a concrete example of a matrix $M$ such that:

1. $M\in Mat_{n}(R)\cap GL_{n}(R^{´})$
2. $h(M)\in GL_{n}(\mathbb{Z})$
3. $M\notin GL_{n}(R)$

Answer 1

This is not exactly an answer to the question, but let me share this anyway. I would like to prove that for an abelian $G$ such $M$ does not exist. This follows from Kaplansky's unit conjecture for abelian groups.

Suppose $G$ is abelian, then $R$ and $R'$ are commutative. Then conditions on $M$ can be restated in terms of determinants, namely, we have that $\operatorname{det}{M}$ is an invertible element of $\mathbb{Z}\left[\frac{1}{2}\right][G]$ but not of $\mathbb{Z}[G]$, and also $\operatorname{det}{h(M)}=\pm1$. So, $\operatorname{det}{M}$ is an invertible element of $\mathbb{Q}[G]\supset \mathbb{Z}\left[\frac{1}{2}\right][G]$, then by Kaplansky's unit conjecture, which is in general false but true for abelian groups, we have $\operatorname{det}{M}=kg$ for some $g\in G$ and $k\in\mathbb{Q}$. Then $k$ is an integer and, since $h$ is a homomorphism, we have $k=kh(g)=h(\operatorname{det}{M})=\operatorname{det}{h(M)}=\pm1$. Thus, $k=\pm1$ and $\operatorname{det}{M}$ is an invertible element of $\mathbb{Z}[G]$, which is a contradiction.