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A number theorist I know (who studies Galois representations) raised a question recently:

Which finite groups can have an irreducible character of degree at least 2 having only $n=2, 3$, or 4 classes where the character takes nonzero values?

He has learned about a few special examples involving nonabelian groups which are very close to being abelian. My limited intuition about the question, based on finite groups of Lie type, suggests that the sort of group he is looking for will be far from simple. But maybe there is no reasonable characterization of these groups for a given $n \leq 4$? In any case, there should be some relevant literature out there. The question itself belongs to a familiar genre in finite group theory: What does the character table tell me about the structure of a group?

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Do you mean the number of group elements at which there are nonzero values, or the number of conjugacy classes of group elements at which there are nonzero values, or the actual number of distinct values taken? In the former case, there are some obvious restrictions, while the restrictions in the last case are not so clear. –  Vipul Naik May 4 '10 at 21:49
@Vipul: Question edited a little. Thanks! –  Jim Humphreys May 4 '10 at 21:58
From what you say you might well know these examples already, but the extra-special groups of order p^{2n+1} have characters of order p^n which are zero outside the centre, and the centre has order p. So for p=2,3 you get examples (which you might well know about already). Disclaimer: I learnt about the existence of extra-special groups from this question here mathoverflow.net/questions/21071/… and the statement I just made about characters was cribbed from Wikipedia. –  Kevin Buzzard May 4 '10 at 22:04
Sorry, my answer was for a different question. You have a single character that is mostly zero, and want to know the group structure? –  Jack Schmidt May 4 '10 at 22:07
I guess the question seems strange to me. A5 has an irreducible ordinary character that is nonzero only on 3 classes, but it only has 5 class to begin with, so this is not surprising. –  Jack Schmidt May 4 '10 at 22:12
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2 Answers

up vote 6 down vote accepted

For the case of n=2 see Gagola's paper "Characters vanishing on all but two conjugacy classes" MR0721927. Some improvements on Gagola's results can be found in Kuisch and van der Waall's papers "Homogeneous character induction [I and II]" MR1172440/MR1302857 and in my paper "Groups with a Character of Large Degree".

See Theorem 4.1 in my paper for what I think is the strongest result in this direction (as far as I know):

Gagola showed that the character of $V$ vanishes on all but one nontrivial conjugacy class if and only if there exists a subgroup $N$ of $G$ such that $N$ acts trivially on each simple $\mathbb{C}[G]$-module except $V$.

Let $G$ be a group of order $n$ and $V$ a simple $\mathbb{C}[G]$-module of dimension $d$. Define $e$ such that $n = d(d+e)$. Assume that there exists a normal subgroup $N \neq > \{1\}$ such that $N$ acts trivially on each simple $\mathbb{C}[G]$-module other than $V$. Let $x$ be a nontrivial element of $N$ and $C$ be the centralizer of $x$ in $G$. Then there exist a prime number $p$, a positive integer $k$ and a non-negative integer $m$ such that:

  • $N$ is elementary abelian of order $p^k$,
  • $C$ has order $p^k e^2$, and $d = e(p^k-1)$, and $n = e^2 p^k (p^k-1)$,
  • $C$ is a Sylow $p$-subgroup of $G$ and $e = p^m$,
  • if $H$ is any group such that $N \subseteq H \subseteq C$ and $|H| > > p^{k+m}$, then $N \subseteq [H,H]$.

The above conditions are also sufficient, see Thm 4.3

For n bigger than 2 there may be some results in MR1487363 Chillag "On zeroes of characters of finite groups."

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Thanks, this looks very helpful and will be passed along. I suspected some literature must exist on questions like this. –  Jim Humphreys May 5 '10 at 12:03
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For n=2, your question is addressed in:

Gagola, Stephen M., Jr. "Characters vanishing on all but two conjugacy classes." Pacific J. Math. 109 (1983), no. 2, 363–385. MR721927 euclid.pjm/1102720107

In it, he shows that if such a nearly-zero character exists, it is unique, and the unique faithful irreducible character of G (similar to the extra-special 2 and 3 groups mentioned by Kevin Buzzard). A doubly transitive Frobenius group has such a character, but the non-solvable doubly transitive Frobenius groups are in short supply (five or so). The possible forms of more general non-solvable examples are restricted in Theorem 5.6 (basically SL2).

Berkovich and Zhmud have results for more general cases. See chapter 27 of their book on character theory (volume 2) which answers a broader question for n=2 and n=3:

Which groups have all irreducible non-linear characters taking on only n distinct values?

The list of groups is fairly short, but I haven't had time to verify it; they are all far from simple, usually with a normal Sylow and nilpotent quotient. They have a result for n=4 too, on p. 239, but it is much less complete.

Since your character is nonzero except on n classes, it can take on at most n nonzero values. It is fairly likely to be faithful, since every class contained in the kernel is a wasted class that cannot be zero. This allows some of the lemmas to be used. If Gagola's case is indicative, then the "all" versus "one" will not actually be a huge difference, at least mod the kernel of your single character.

Here are the paper references:

Berkovich, Yakov; Chillag, David; Zhmud, Emmanuel. "Finite groups in which all nonlinear irreducible characters have three distinct values." Houston J. Math. 21 (1995), no. 1, 17–28. MR1331241

Zhmudʹ, È. M. "On finite groups, all of whose irreducible characters take at most two nonzero values." Ukraïn. Mat. Zh. 47 (1995), no. 8, 1144–1148;translation in Ukrainian Math. J. 47 (1995), no. 8, 1308–1313 (1996) MR1367729 DOI:10.1007/BF01057720

Berkovich and Zhmud also pose an exercise that if all the irreducible characters of degree coprime to p take on only 3 values, then the group is p-solvable. The same is true if "coprime to" is replaced by "divisible by". This is page 238.

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Thanks for the further references, which overlap Noah's somewhat but include other interesting directions. –  Jim Humphreys May 5 '10 at 11:54
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