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Suppose we are working over $\mathbf{Z}[G]$ where $G$ is finite. Suppose further we have two representations $\rho$ and $\rho^\prime$ such that $\rho^\prime=(\rho)^T$. Can we say that these two representations are equivalent?

I know that if we were working over $\mathbf{C}$ then we could consider characters since these two representations have the same character and are therefore equivalent, but does this method survive the transition to integral representation?

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  • $\begingroup$ Do you additionally want $G$ to be abelian? If $G$ is nonabelian, then $\rho '(ab)=\rho '(b)\rho'(a)$ may not satisfy the correct multiplicative rule. $\endgroup$ – rschwieb Aug 19 '14 at 13:27
  • $\begingroup$ I didn't really want to impose any conditions on $G$ beyond finiteness. $\endgroup$ – Sam Williams Aug 19 '14 at 13:38
  • $\begingroup$ But... how will it be a representation if the multiplicative property isn't satisfied? Or it could be that I've forgotten there's another meaning to the transpose in representation theory. Is it not just the transpose of matrices representing $G$ via $\rho$? $\endgroup$ – rschwieb Aug 19 '14 at 13:39
  • $\begingroup$ Do you know of any conditions I could impose on $G$ that aren't quite as extreme as that of commutativity? $\endgroup$ – Sam Williams Aug 19 '14 at 13:39
  • $\begingroup$ Yes that's right. Essentially I have been trying to show two representations are equivalent apart from the actual formal sum calculations which look extremely messy. I noticed that the one is the inverse of the transpose of the other. Clearly the inverse will give an equivalent representation since that is just a succession of elementary row and column operations, but I am now left with the question of whether taking the transpose will still give me an equivalent representation. $\endgroup$ – Sam Williams Aug 19 '14 at 13:41

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