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I am aware that, given a group, there is no simple general procedure to construct the character table of the group (over complex numbers). However, for specific groups, we could use helpful additional information regarding the characters, (apart from orthogonality relations, etc.) to construct the character table. For instance, we know that the characters of a symmetric group are all integers.

Now suppose we have the group $S_5$, the symmetric group on 5 elements. Suppose we are given only the degrees of the irreducible representations/characters, namely $1,1,4,4,5,5,6$. Can this alone (apart from general results on characters) be used to derive the entire character table of $S_5$?

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How are you defining 'simple' –  Ethan Aug 4 '13 at 3:47
    
@Ethan: By simple, I mean finding the character table using the equations determined by elementary character-theoretic results like orthogonality relations. –  BharatRam Aug 4 '13 at 4:17
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An easy (read: that can be implemented in a computer) way to compute the character table of a finite group is Burnside's algorithm, which translates the problem of computing the character table to the problem of finding eigenvalues/eigenvectors of matrices.

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This algorithm is essentially what is used in the so-called Dixon-Schneider algorithm, which is used as a default method by GAP, for example. The principal idea contributed by Dixon was to do the eigenvector computations in a finite field (rather than over a number field or using floating point arithmetic) and then to lift the result back to a number a field. Schneider contributed a lot of heuristics to avoid having to do too much conjugacy testing in the group. –  Derek Holt Aug 4 '13 at 17:24
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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.

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I am aware that, given a group, there is no simple general procedure to construct the character table of the group (over complex numbers).

Depends on what you mean by "simple." There are algorithms which will spit out the character table of a finite group.

Now suppose we have the group $S_5$, the symmetric group on 5 elements. Suppose we are given only the degrees of the irreducible representations/characters, namely 1,1,4,4,5,5,6. Can this alone (apart from general results on characters) be used to derive the entire character table of $S_5$?

Just being given that it's $S_5$ is enough to derive the character table, so this question doesn't make sense.

However, let's say you're given the dimensions of the representations, but no other information about the group. That's not enough information to identify the group or reconstruct the character table in general. For instance, suppose I showed you that a group has four irreps, each one-dimensional. Well, now you know it's an abelian group of order four, but which one? They have different character tables.

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Hi. I didn't mean it in a very precise algorithmic or information-theoretic way. What I meant was that can we derive the character table using elementary information about the group (its conjugacy class structure, subgroups) and simple theorems from character theory (orthogonality relations), without using any involved results. I admit, its debatable what "elementary" or "involved" means, but I hope the rough picture is clear. –  BharatRam Aug 4 '13 at 4:48
    
Well, as I've shown, knowing the size of the conjugacy classes and dimensions of the irreps isn't enough information. I can't give an example off the top of my head, but I'm sure that throwing in the graph of the subgroup lattice still wouldn't be enough. –  Daniel McLaury Aug 4 '13 at 4:53
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