Could you give me an example of two non-isomorphic groups with the same complex character table?
Two Non-Isomorphic Groups with Same Complex Character Table
charactersgroup-theoryrepresentation-theory
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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.
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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.
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If $G$ is a finite group whose irreducible characters have degrees 1,1,4,4,5,5,6 then $G$ is isomorphic to $S_5$.
From the character degrees we immediately get that $G/[G,G] \cong C_2$ since there are exactly two irreducible characters of degree 1, and that $|G|=1^2 + 1^2 + 4^2 + 4^2 + 5^2 +5^2 +6^2 = 120$. We also know that such a group has only 7 conjugacy classes.
If you have classified the groups of order 120, then you know there are only 3 groups with $G/[G,G] \cong C_2$ and only one of those has 7 conjugacy classes, namely $G=S_5$.
If not, you can already extract a lot of information. From the character degrees we know that $G$ is not of the form $H \times C_2$ (only one copy of $6$). We know the index of the Sylow 5-subgroup is either 1 or 6, but if it is 1, then the character degree 5 is not possible (standard result from Isaacs's textbook, 6.15), so we get a group with 6 Sylow 5-subgroups. We know the focal subgroup for $p=2$ is index 2, and consulting our table of groups of order 8, we get the Sylow 2-subgroup is $C_2 \times C_2 \times C_2$ with a direct factor (no!) or $D_8$ with PGL fusion ($S_5$ is PGL(2,5)). So we already get the 2-local and 5-local structure.
In general though the character degrees don't have to tell you much about the group. What they do tell you about the group is an area of active research. The state of affairs in the late 1960s is summarized in chapter 12 Isaacs's textbook, and I believe many of his more recent papers have more up to date summaries.
Best Answer
The dihedral group and the quaternion group.