Consider $\mathbf{Z}G$, $G$ finite. If the characters of two $\mathbf{Z}G$-modules are equal, does it follow that the modules are isomorphic?

Question

So I have recently started to delve into integral representation theory and I was wondering if a particularly useful theorem survives the transition to integral rep theory. Basically, suppose we have two $\mathbf{Z}G$ modules $M$ and $N$ such that their representations are isomorphic. Clearly this means they have equal characters. What I am wondering is whether this implies $M\cong N$, and if so, where I can find such a proof. I am interested in this since obviously this holds over fields of characteristic zero which are algebraically closed. Any help in this matter will be greatly appreciated. Thanks in advance!

Answer 1

The answer is no. One example to keep in mind is the 2 dimensional representation of $\mathbb{Z}/p$ over $\mathbb{F}_p$ where 1 acts by the matrix $$\left( \begin{array}{ccc} 1 & 1 \\ 0 & 1\end{array} \right)$$

This representation has trivial character.

Answer 2

If you want an example where the modules are free over $\mathbb{Z}$, then let $G=\langle g\rangle$ be a cyclic group of order $2$, let $M=\mathbb{Z}G$ be the regular $\mathbb{Z}G$-module, and let $N=U\oplus V$ be a direct sum of two copies of $\mathbb{Z}$, where $g$ acts trivially on $U$ and by multiplication by $-1$ on $V$.