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.

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    $\begingroup$ Note that the first property you mention (images are either traceless or multiples or the identity) means that the center of the character and the vanishing-off subgroup coincide, and both of these subgroups have been studied a fair bit. $\endgroup$ Feb 23, 2011 at 12:52
  • $\begingroup$ @TobiasKildetoft, what is "the center of the character" (or does it mean "… of the group")? $\endgroup$
    – LSpice
    Oct 31, 2018 at 21:17
  • $\begingroup$ @LSpice it is the set of all g in G for which phi(g) has modulus 1 (see Frieder Ladisch's answer below). This is terminology going back to old-school character theory of finite groups $\endgroup$
    – Yemon Choi
    Nov 1, 2018 at 1:45
  • $\begingroup$ Thanks! For clickability, here's @FriederLadisch's answer. $\endgroup$
    – LSpice
    Nov 1, 2018 at 1:59

2 Answers 2


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).

  • $\begingroup$ Many thanks for this detailed answer. I need to take some time to digest this, but it looks to be just the kind of thing I was after. $\endgroup$
    – Yemon Choi
    Feb 24, 2011 at 19:57
  • $\begingroup$ The examples you give, and the proof that 1-minimal groups are nilpotent, are a great help. (With hindsight I should really have thought of looking at extra-special groups.) I've taken the liberty of passing on your comments to my coauthors. Would you mind if we cited the argument you give here, attributed to you of course? $\endgroup$
    – Yemon Choi
    Feb 26, 2011 at 0:12
  • $\begingroup$ Actually... I am slightly embarrassed to admit that on a closer reading there is a step I don't follow. Namely, you have a minimal nontrivial normal subgroup $N$, and a $1$-minimal character $\chi$ whose restriction to $N$ is not identically $1$. You then say that $N\cap Z(\chi)$ cannot be trivial, because if it were trivial then the trivial character of $N$ would contain a copy of $\chi_N$. Could you please explain why this cannot be the case, since $\chi_N$ could be very far from being irreducible? $\endgroup$
    – Yemon Choi
    Feb 26, 2011 at 22:48
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    $\begingroup$ @Yemon: No, I would not mind if you use this. Concerning the other question: It is a theorem of A. H. Clifford that all irreducible constituents of $\chi_N$ are conjugate in $G$, when $\chi$ is irreducible and $N$ a normal subgroup of $G$ (in Isaacs' book Character Theory of Finite Groups, this is Theorem 6.2). So if $\chi_N$ would contain $1_N$, then we would in fact have $\chi_N = \chi(1)1_N$, since the trivial character of $N$ is only conjugate with itself. $\endgroup$ Feb 27, 2011 at 21:29
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    $\begingroup$ If $\chi$ is "1-minimal" in your terminology, one sometimes says that $\chi$ is "fully ramified" over $Z(\chi)$. This terminology is actually used in a wider context, namely, if $N$ is normal in $G$ and $\chi_N = e\theta$ with $\theta\in \text{Irr}(N)$, and $e^2= |G:N|$, one says that $\chi$ is fully ramified over $N$ and $\theta$ is fully ramified in $G$. This situation has been studied by Dade, Isaacs and others, since it occurs in the character theory of finite solvable groups. $\endgroup$ May 18, 2011 at 16:15

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.

  • $\begingroup$ Many thanks! I will pass this information on to my coauthors. I am also updating the original question to include a link to the preprint that has resulted from the work mentioned there. $\endgroup$
    – Yemon Choi
    Jan 6, 2012 at 2:33

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