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For a finitely presented group, does there always exist a nontrivial finite dimensional unitary representation? If two finitely presented groups have the same set of finite dimensional unitary representations, are they necessarily isomorphic?

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What does it mean for the set of finite-dimensional unitary representations to be the "same"? – Qiaochu Yuan Dec 2 '10 at 12:59

No, a finitely presented simple group does not have any non-trivial finite dimensional matrix representation because every finitely generated matrix group is residually finite. There are many non-isomorphic finitely presented simple groups: Thompson groups $T$ and $V$, Burger-Mozes groups and others. They all have the same fin. dim. unitary representations (trivial).

Update A more interesting question is whether two non-isomorphic residually finite groups can have the same finite dimensional matrix (unitary) representations. The answer, I think, is "yes", and is given by the class of groups, found by Nekrashevych in Trans. Amer. Math. Soc. 362 (2010), no. 1, 389–398. Any two groups in his class have the same finite quotients and every linear representation of any of these groups is finite (the latter result follows from the ``branch" property).

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Well, if you have two nonisomorphic groups neither of which has finite-dimensional matrix representations, then certainly the answer to your updated question is yes! – Theo Johnson-Freyd Dec 2 '10 at 19:03
The groups are residually finite so they have lots of finite dimensional unitary representations. I have answered the updated question also. But it is much less trivial than the original question. – Mark Sapir Dec 2 '10 at 19:43

One can even find a nice criterion: By Malcev's theorem, every finitely generated linear group is residually finite. Conversely, a finite group clearly admits a nontrivial unitary representation. So a finitely generated group admits a nontrivial unitary representation if and only if it has a nontrivial finite quotient.

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