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As is already known, groups $D_8$ and $Q_8$ have equivalent character tables but $\mathbb{Q}D_8 \not= \mathbb{Q}Q_8$. Does anyone know if given an isomorphism between group algebras the character tables are equivalent for finite groups? In order words, does the following statement hold?

$KG \cong KH \implies T_G \cong T_H$,

where $T_G$ is the character table for G and $T_H$ for H.

Thanks for the help.

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    $\begingroup$ Characters are representation dependent. Easiest non-trivial example represention of SU(2) and SO(3) with identical abstract Lie algebras. and their infinite series of irreducible representations by operators in Hilbert spaces $\endgroup$
    – Roland F
    Apr 2 at 17:19
  • $\begingroup$ Sorry, I'm talking about finite groups, I forgot to specify. I'm editing the question $\endgroup$
    – ayphyros
    Apr 2 at 17:21
  • $\begingroup$ Then you ought to add the proper tag, @ayphyros $\endgroup$
    – Kan't
    Apr 2 at 17:27
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    $\begingroup$ @citadel sorry, I did not know it existed. Added. Thanks :) $\endgroup$
    – ayphyros
    Apr 2 at 17:32

1 Answer 1

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Over an algebraically closed field in characteristic zero (think $\mathbb{C}$ if you like), isomorphism of group algebras depends solely on the degrees of the irreducible representations.

This means that the group algebras of the dihedral group of order 12, and of the semidirect product $C_3:C_4$ will be isomorphic, but the character table of the former one has only rational entries, by the latter one involves the complex root of -1, so the tables cannot be isomorphic (whatever the definition).

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