characters on a finite group with `extremal' behaviour The following question is a bit technical, and I haven't got to grips with it enough to be able to present it as a well-focused question. However, my hope is that the collective group-theoretic expertise on MO may spot something in there that follows from standard results, or which is likely to be intractable by way of connection to open problems.
Let $G$ be a finite group, equipped with normalized counting measure. If $\chi$ is an irreducible character on $G$ then standard theory/calculations show that it has $\ell^2$-norm equal to $1$; it follows easily from this that the $\ell^1$-norm of $\chi$ is $\geq d^{-1}$ where $d$ is the degree of $\chi$. If equality actually holds (which, as I'll discuss below, seems to be very restrictive) then let's say that $\chi$ is $1$-minimal (non-standard terminology).
$1$-minimal characters of degree $>1$ give rise to some interesting ideals in the centre of (algebras built from) $\ell^1(G)$, which I and some colleagues have been looking at. We have run into the following question which seems to lie out of our main expertise.
Question 1. for which groups $G$ are all the irreducible characters $1$-minimal?
Call such a group $1$-minimal, for want of a better term at present.
Every abelian group is $1$-minimal, as is the dihedral group of order 8 (to my initial surprise). It is not hard to show that products of $1$-minimal groups also have the property.
Here is the small amount I can show so far. A little calculation shows that if $\pi$ is an irrep of $G$, then its trace $\chi$ is a $1$-minimal character if and only if $\pi(x)$ is either traceless or a scalar multiple of the identity. Writing $d$ for the degree of $\pi$, it follows that the support of $\chi$ is a normal subgroup $K$ of $G$, containing $Z(G)$, with the index of $K$ in $G$ being $d^2$. This already seems to preclude many groups from being $1$-minimal.
The above shows that $d\chi$ is induced from some 1-dimensional representation of $K$, but I haven't succeeded in using this to get much leverage. Other partial results: if $\pi$, $\chi$, $d$, $K$ are as above, and every non-linear character of $G$ has degree $\geq d$, then one can play around with the conjugation representation associated to $\pi$ to show that $G/K$ is abelian. Unfortunately, restricting attention to cases where $d$ is small makes it harder to get control on $K$...
So far, the only examples of $1$-minimal groups that I have found are products of abelian groups with copies of $D_4$. All these examples are solvable and have non-trivial centre, which leads me to some more focused questions.
Question 2. Can we characterize those solvable, finite groups which are $1$-minimal?
Question 3. Do there exist $1$-minimal groups with trivial centre?
A final remark/piece of context: some foraging via Google and MathSciNet suggests that this problem might be connected with the old problem of characterizing the finite groups of "central type", namely those groups $G$ for which there is an irreducible character of degree $d$ where $d^2= | G: Z(G) |$ ( we always have $d^2\leq |G:Z(G)|$ for any group). However, it is not clear to me that the present question is either contained in, or contains, this older one.

Update 5th Jan. 2012: Thanks to the arguments given below by F. Ladisch and M. Isaacs, we have the following: every $1$-minimal group is nilpotent (so that Question 3 has a negative answer if we exclude the trivial group) and every nilpotent group of class $2$ is $1$-minimal.
For an explanation of why I and my colleagues were interested in the condition of $1$-minimality, see Theorem 4.2 of our preprint 1110.6683. Note for the record that parts of the paper form part of the PhD thesis of the first author, which is at time of writing still in progress.
 A: All finite groups that are nilpotent with nilpotency class at most 2 are 1-minimal in the sense of this question. Let $\chi \in {\rm Irr}(G)$ and write $Z = {\bf Z}(\chi)$. By Theorem 2.31 of my character theory book, $|G:Z| = \chi(1)^2$ if $G/Z$ is abelian. This condition always holds if $G$ has nilpotence class $2$, because in that case, $G/{\bf Z}(G)$ is abelian, and we certainly have ${\bf Z}(G) \subseteq Z$. By Corollary 2.30, it follows that $\chi$ vanishes on $G - Z$, and we know that this condition is equivalent to saying that $\chi$ is 1-minimal.
A: Dear Yemon Choi, $\DeclareMathOperator{\Z}{\mathbf{Z}}\DeclareMathOperator{\Irr}{Irr}$
here is what I know about this question.
First, two classes of examples: Extraspecial $p$-groups are $1$-minimal groups, as are the following groups: The automorphism group of $C_{p^n}$ ($p$ prime) has order $(p-1)p^n$. Let $A$ be the Sylow $p$-subgroup of $\operatorname{Aut}(C_{p^{n+1}})$ and let $G$ be the semidirect product of $A$ and $C_{p^{n+1}}$. Then $G$ is a $1$-minimal group of order $p^{2n+1}$, with characterdegrees $1$, $p$, $\dotsc$, $p^n$ occuring.
Second, answering your Question 3, the following is true:

Every $1$-minimal group is nilpotent.

As a consequence, every $1$-minimal group is a direct product of $1$-minimal $p$-groups.
I use the standard notation $$\Z(\chi)=\{ g\in G \mid |\chi(g)| = \chi(1) \} $$
to denote the center of a character $\chi$, which is the set of group elements acting as scalar multiple of the identity in the underlying representation. Moreover, for $\chi\in \Irr(G)$ it is a standard result that $\Z(\chi)/\ker \chi = \Z(G/\ker\chi )$.
In particular, in a $1$-minimal group, for every $\chi\in \Irr(G)$ the factor group $G/\ker\chi$ is of central type.
Now assume that $G$ is $1$-minimal. We only need to show that $\Z(G)>1$, because then induction will yield that $G/\Z(G)$ is nilpotent (factor groups of $1$-minimal groups being $1$-minimal), and so $G$ itself is nilpotent.
Thus, let $ N$ be a minimal nontrivial normal subgroup of $G$. We show that $N\subseteq \Z(G)$. There is some $\chi\in \Irr(G)$ such that $N\not\subseteq \ker\chi =:K$. By minimality, either $N \subseteq \Z(\chi)$ or $N\cap \Z(\chi) =1$.
If the latter is the case, then $\chi$ vanishes on $N\setminus 1$. But then $(\chi_N, 1_N)>0$ which is only possible when $N$ is contained in the kernel of $\chi$, contradiction. Thus $N\leq \Z(\chi)$. It follows that $[N,G] \subseteq K$. But since $N$ is normal in $G$, we also have $[N,G]\subseteq N$, and thus $[N,G] \subseteq N\cap K=1$, where the last equality follows from the minimality of $N$. This means that $N\leq \Z(G)$ as desired.
A further remark that might be useful: Given $\chi\in \Irr(G)$, one can define a "bilinear form", which is only defined for pairs $(g,h)\in G\times G$ such that their commutator $[g,h]=g^{-1}h^{-1}gh$ is in $\Z(\chi)$, namely $a(g,h)= \chi([g,h])/\chi(1)$. (If $G/\Z(\chi)$ is abelian, this is defined on all of $G\times G$.) Then it is known that $\chi$ is of central type ($1$-minimal in your terminology) if and only if this form is non-degenerate as a form on $G/\Z(\chi)$ in the sense that for every $g\in G\setminus \Z(\chi)$ there is $h\in G$ such that $[g,h]\in \Z(\chi)$, but $a(g,h)\neq 1$. This follows from a result of Gallagher. The form was introduced by Isaacs (Characters of Solvable and Symplectic Groups, 1973).
