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This question was just raised by a colleague (who shall for the moment remain anonymous). It may or may not have a reasonable answer.

For which finite nonabelian groups $G$ do all irreducible complex characters of degree $>1$ have at most one strictly positive real value (namely the degree)?

One small example is the alternating group $A_4$. This could be viewed as a sort of degenerate Chevalley group, whose Steinberg character of degree $3$ has other values $-1, 0, 0$. But for most finite groups of Lie type over a finite field of order $q$, the Steinberg character of degree a power of $q$ will take on more than one positive integral value. While an enumeration of finite groups having the stated property may well be out of reach, it would be of interest to know:

Are there infinitely many groups $G$ with the special property stated above?

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Well, for the first question, all abelian groups work so the answer to the second question is yes :) –  Mariano Suárez-Alvarez Feb 22 '11 at 20:40
    
After thinking for only a few minutes about this, if $G$ has even order then all elements of order 2 (even if they are in different conjugacy classes) must lie in a single normal subgroup of $G$ (this applies to your example of $A_4$, and it seems the condition might even strengthen to all elements of even order). The proof is by considering column orthogonality between a class of elements of order 2 with the identity class; the elements of order 2 must have character values all negative or equal to the degree whereupon the elements of order 2 all act trivially in the quotient. –  ARupinski Feb 22 '11 at 21:46
    
@ARUpinski: simple groups of non-prime order have even order, and their only normal subgroups is the total one... therefore the claim in your first sentence can only be true if when you say "[they] must lie in a single normal subgroup of $G$" you include the possibility that the normal subgroup be $G$ itself. But then, the statement is not very interesting. –  Mariano Suárez-Alvarez Feb 22 '11 at 21:51
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There are indeed infinitely many such groups, for instance $G=C_p\rtimes C_{p-1}$ for any odd prime $p$. It has only one irreducible representation of dimension $>1$, with character $(p-1,0,0,...,0,-1)$. –  Tim Dokchitser Feb 23 '11 at 0:24
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@Jim: Simple groups other than prime cyclic ones cannot satisfy the condition by column orthogonality between the identity and an element of order 2. However, I must admit that the more I think about it, I am worried there is a major hole in my quick proof about only having one conjugacy class of order 2 elements, even in the perfect case (so much for doing math at home instead of around other math people) although I don't have any counterexamples. The comment about 2-groups still holds. –  ARupinski Feb 23 '11 at 18:02

2 Answers 2

up vote 5 down vote accepted

Here are some classes of such groups I can think of:

  • extraspecial $p$-groups (and probably some other $p$-groups)
  • metacyclic groups, where $\lvert G: G'\rvert$ has odd order
  • the groups $AGL(1,q)$ (this subsumes Tim Dokchitser's examples and the $A_4$ example). More generally, suppose a group $H$ with few positive real values acts sharply transitive on the nonzero elements of a finite vector space $V$. Then the semidirect product $HV$ has also only characters with at most one strictly positive value. An example is $Q_8$ acting on $C_3\times C_3$.

Here is an argument, that there is probably no simple group of this type rules out some groups: If a group $G$ acts $2$-transitive on some set, then the permutation character $\pi$ has the form $\pi= 1+\chi$ with $\chi \in \operatorname{Irr}(G)$. By assumption, if $g$ acts nontrivially, then $\chi(g)\leq 0$. This implies that every element not in the kernel $K$ of the action has at most one fixpoint. But then $G/K$ is a Frobenius group, the elements with no fixpoints form the Frobenius kernel.
While I'm not an expert for finite simple groups, I suppose this argument rules out many (all?) finite simple groups, namely all that admit a $2$-transitive permutation representation.
Since the property in the question is inherited by factor groups, this would also yield that there are no perfect groups with that property.

EDIT: Thanks to Jack for putting that right. Anyway, simple groups are ruled out by ARupinksi's comment.

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2-transitive reps are sort of rare. I put a list from Kantor in an answer at: math.stackexchange.com/questions/16954/… Basically it is alternating, psl, a few low rank groups, and some sporadic examples (as well as the affine ones like your AGL, which are never simple). –  Jack Schmidt Feb 23 '11 at 18:28

Here is an argument that resticts the structure of such groups of even order, which expands on ARupinksi's comments above (inspiration came from a nice argument in this paper).
Namely, suppose that $1\neq t$ is a real element in $G$, where $G$ is as in the question. Form the following normal subgroups: $$ K = \langle t \rangle^G = \bigcap_{\chi(1)=\chi(t)} \ker \chi \quad \text{and} \quad L = \bigcap_{\chi(t) \neq 0} \ker \chi < K , $$ where the intersections run over irreducible characters $\chi$ of $G$. Then we have $$ t^G = K \setminus L,$$ that is, the elements in $K\setminus L$ are all conjugate in $G$ to $t$. In particular, it turns out that $t$ is rational in the sense that it is conjugate to every generator of $\langle t \rangle$, which is equivalent to every character having rational value at $t$.
Why is this true? Well, column orthogonality yields $\DeclareMathOperator{\Irr}{Irr}$ $$ 0 = \sum_{\chi \in \Irr G} \chi(t)\chi(1) = |G:K| + \sum_{\chi(t)<\chi(1)}\chi(t)\chi(1). $$ Now let $y\in K$ and suppose that $t$ and $y$ are not conjugate. Then plugging in $y$ instead of $1$ in the last formula, we see that $$ \sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(y) = |G:K| =\sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(1) .$$ Since $-\chi(t)>0$ and $|\chi(y)|\leq\chi(1)$, it follows that $y$ is in the kernel of every irr char $\chi$ with $\chi(t)< 0$, that is, $y\in L$. Thus $K\setminus L = t^G$.
(Added later:) If $t$ is an involution, then it follows that $x^t=x^{-1}$ for all $x\in L$, in particular $L$ must be abelian, and elements of $L$ are real.

The "dual" argument (exchanging the roles of characters and conjugacy classes) shows the following: Suppose $1\neq \chi$ is real valued, and let $V= \operatorname{\mathbf{V}}(\chi)$ be the vanishing-off group of $\chi$, generated by all group elements on which $\chi$ is non-zero. Then $$ \Irr( G/\ker \chi ) = \{ \chi \} \cup \Irr(G/V). $$ From this it follows easily that $V/\ker\chi$ is a conjugacy class of $G/\ker\chi$ and that the only value of $\chi$ besides $0$ and $\chi(1)$ is $-\chi(1)/(|V/\ker\chi|-1)$. (In particular, any real character is rational.) Groups with such an character have been studied by Zhmud, where more information can be found.
I suppose there is also literature on groups having normal subgroups $L\subset K$ such that $K\setminus L$ is a conjugacy class of $G$. The notion of a Camina pair/group seems to be related (see this paper and papers that refer to it).

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Thanks for adding so much detail, which is important for those of us who aren't close to this kind of literature. (For example, it took me a while to sort out a short comment by Tim earlier about semidirect products of cyclic groups in a situation where there is more than one way to form a nontrivial semidirect product.) My experience has been that people who know only the most basic character theory and special examples tend to assume that all questions can be answered smoothly. –  Jim Humphreys Feb 24 '11 at 23:02
    
The bottom line is, I think, that you can narrow down the possibilities for such groups a lot, but a satisfying classification might be difficult. –  Frieder Ladisch Feb 25 '11 at 0:50

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