Let $G$ be a finite group. I'm looking for two nonisomorphic $\mathbb{R}G$-modules $U$ and $V$ with the same characters. Could someone provide an example?

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    $\begingroup$ If you mean finite-dimensional modules, then fat chance. A finite-dimensional representation of a finite group over a field of characteristic $0$ is uniquely determined (up to isomorphism) by its character. This is proven in Corollary 4.2.4 of Pavel Etingof et al., Introduction to representation theory, AMS 2011 when the field is algebraically closed. When the field is not algebraically close, you need to additionally use the fact that two representations that become isomorphic over the algebraic closure must already ... $\endgroup$ – darij grinberg Jan 23 '18 at 20:24
  • $\begingroup$ ... be isomorphic over the original field (see math.stackexchange.com/questions/243669/… ). $\endgroup$ – darij grinberg Jan 23 '18 at 20:24
  • $\begingroup$ @darijgrinberg is right about what happens over a field of char. 0, but I would like to add that this question becomes very interesting if $\mathbb{R}$ is replaced by $\mathbb{F}_p$ or $\mathbb{Z}$. $\endgroup$ – Steve D Jan 23 '18 at 21:00
  • $\begingroup$ @SteveD is that known in those cases? $\endgroup$ – Alexander Gruber Jan 25 '18 at 4:21
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    $\begingroup$ @SteveD Ahh! I looked around and found another question which helps out. $\endgroup$ – Alexander Gruber Jan 25 '18 at 5:46

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