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Suppose two groups have the same character table of complex representations. Also, all the entries in this character table have absolute value at most $1$. Does this imply that the two groups are isomorphic?

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Are those finite groups you're asking about? –  tomasz Jun 15 '12 at 19:56
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Do character tables even make sense if the groups are not finite? –  William Jun 15 '12 at 19:58
    
there is no information about groups if they are finite or not –  srp Jun 15 '12 at 20:04
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William: they might still make sense for compact groups. –  tomasz Jun 15 '12 at 20:10
    
You can think of the character table as a pairing between two sets, the irreducible finite-dimensional representations and the conjugacy classes, that gives a complex number. For two arbitrary groups it still makes sense to ask for a bijection between the conjugacy classes and irreducible representations of the two groups that allows us to identify the two pairings. –  Qiaochu Yuan Jun 15 '12 at 20:23

1 Answer 1

If you're asking about finite groups, then yes.

All entries have absolute value at most 1, so in particular all irreducible representations are one-dimensinal, so the groups in question are abelian. Thus we can apply the fundamental theorem of finitely generated abelian groups to them to decompose them into direct sums of cyclic groups.

==edit==

As per Jyrki Lahtonen's suggestion, there's a simpler way to finish the proof: a finite abelian group is isomorphic to its dual (as in the character group), which can be shown using the decomposition inductively: for cyclic groups it is trivial, and the dual of a direct sum is a direct sum of the duals (which is again not very hard to see). From character table we can deduce the character group, so we're done.

If I'm not mistaken, investigating the character group is also beneficial in the general (locally compact) case. The Pontryagin/van Kampen duality theorem states that for any locally compact abelian group, there is a natural (topological) isomorphism between it and its double dual, defined by the $\varphi(x)(\chi)=\chi(x)$ formula. On the other hand, we can read the character group, and consequently the bidual from the character table.

The duality theorem is quite strong a tool, though. You can find the details in e.g. Hewit, Ross: Abstract harmonic analysis, vol. 1.

==/edit==

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+1 Finite abelian groups are isomorphic to their groups of complex characters: this holds trivially for cyclic groups, and taking the dual commutes with direct products. –  Jyrki Lahtonen Jun 15 '12 at 21:28
    
what is the relation between the decomposition of the group and orders of irreducible representaions of the group? –  srp Jun 17 '12 at 8:45
    
@srp: I don't think it's anything simple, but I'm pretty sure it's 1-to-1 (it's just combinatorics involving representations of order p^k for prime p and varying k). Anyway, Jyrki Lahtonen's comment is simpler and more elegant. I will make an edit to reflect that. –  tomasz Jun 18 '12 at 17:16

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