# 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?

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!

• I am a little unsure what you mean here. Do you mean you have two $\mathbf{Z}G$-modules with equal characters, and you want to know if this implies that the modules are isomorphic? – Tobias Kildetoft Aug 12 '14 at 13:43
• Yes, that is what I mean. – Sam Williams Aug 12 '14 at 13:50

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 $V$.

• Hi Jeremy, thank you for answering. Perhaps I am being a bit slow but are you supposing this is a counter example to my question? – Sam Williams Aug 12 '14 at 22:20
• @SamWilliams Yes, both modules have the regular character, but they are not isomorphic, since $N$ is a direct sum of two rank $1$ modules but $M$ is not. – Jeremy Rickard Aug 13 '14 at 5:39
• Ah okay, now this makes more sense. Do you know if there is ever a case where this holds? Such as for lattices etc? – Sam Williams Aug 13 '14 at 14:02
• @SamWilliams If by "lattice" you mean a $\mathbb{Z}$-free $\mathbb{Z}G$-module, then the modules in my example are both lattices. – Jeremy Rickard Aug 13 '14 at 15:36
• By lattice I meant a $\mathbf{Z}G$ module whose underlying abelian group is both finitely generated and torsion free. – Sam Williams Aug 13 '14 at 19:51

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.

• Perhaps I m missing your point but that is because we are working over a finite field, rather than the integers which are certainly infinite. – Sam Williams Aug 12 '14 at 22:17
• Working over a finite field is over the integers. ZG modules are just abelian groups with G actions. – jspecter Aug 12 '14 at 22:36
• Note that in Jeremy's answer $M \otimes \mathbb{F}_2$ is exactly the representation of $Z/p$ described in my answer for $p = 2.$ In fact, the fact $M \not\cong N$ can be attributed to exactly this problem at 2 as $M[1/2] \cong N[1/2].$ – jspecter Aug 12 '14 at 22:39
• Thanks, this has cleared up things for somewhat – Sam Williams Aug 13 '14 at 14:02