Are two groups isomorphic if they have the same character table and each $|\chi| \leq 1$? 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?
 A: 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==
