When you browse the character tables of the small finite groups (for example here), you can observe that every zero entry corresponds to the value of an irreducible character $\chi$ on a non-central element $g$ such that the degree $\deg(\chi)$ of $\chi$ and the order $|C_G(g)|$ of the centralizer of $g$ in $G$ are not coprime (i.e. $\gcd(\deg(\chi) , |C_G(g)| ) \neq 1$).

Question 1: Is it true in general?

The reciprocal is false, one counter-example is given by $S_4$ which admits an irreducible character $\chi$ and a non-central element $g$ with $\deg(\chi) = 2$, $|C_G(g)| = 8$ but $\chi(g) = 2$. Moreover, the vanishing of $\chi(g)$ for $\chi$ irreducible ang $g$ non-central is not completely determined by $\deg(\chi)$ and $|C_G(g)|$ because for $G = M_{11}$ there is $\chi$ irreducible, $g_1, g_2$ non-central with $\deg(\chi) = 10$, $|C_G(g_1)| = |C_G(g_2)| = 8$, but $\chi(g_1) = 0$ and $\chi(g_2) = 2$.

Question 2: Why for $\chi$ irreducible, $g$ non-central and $\gcd(\deg(\chi) , |C_G(g)| ) \neq 1$ then "often" $\chi(g)$ vanishes? Is it always true in some specific cases, for example when $\deg(\chi) = |C_G(g)| $?

  • $\begingroup$ Just to parse the logic, you make the following conjecture: If $G$ is a finite group, $\chi$ a character of $G$, and $g \in G$ such that $|C_G(g)|$ and $\deg(\chi)$ are coprime, then $\chi(g) \neq 0$. Did I get it right? $\endgroup$ – Theo Johnson-Freyd Jun 22 '20 at 12:51
  • 2
    $\begingroup$ Dear Sebastien, here are two facts which partially explain your observations: (1) The central character $\frac{|G|\chi(g)}{|C_G(g)|\chi(1)}$ of $\chi$ at the conjugacy class of $g$ is an algebraic integer. When gcd$(\chi(1),|C_G(g)|)\ne1$, this makes it more likely that $\chi(g)=0$. (2) Let $p$ be a prime and suppose that $|G|/\chi(1)$ is coprime to $p$. Then Richard Brauer proved that $\chi(g)=0$ if $p$ divides the order of $g$. Note that in this situation $p$ divides gcd$(\chi(1),|C_G(g)|)$. $\endgroup$ – John Murray Jun 22 '20 at 12:55
  • $\begingroup$ @TheoJohnson-Freyd: This is the statement that Question 1 asks for, but for now it is just an observation on (very) small groups, not a general conjecture (yet). $\endgroup$ – Sebastien Palcoux Jun 22 '20 at 13:27
  • 1
    $\begingroup$ See (3.7) in Isaacs, Character Theory of Finite Groups, Academic Press or G. James, M. Liebeck, Representations and Characters of Groups, Academic Press. $\endgroup$ – John Murray Jun 22 '20 at 13:57
  • 2
    $\begingroup$ Maybe I misunderstand something, but it seems to me that if $\chi$ is the irreducible character of $S_6$ corresponding to the partition $(3,3)$ and $g \in S_6$ is a $6$-cycle, then $\chi(g)=0$ as $(3,3)$ does not have hook shape, $\chi(1)=5$ by the hook-length formula or counting standard Young tableaux or a well-know fact about Catalan numbers, and $C_{S_n}(g)=\langle g \rangle$ has order six. $\endgroup$ – John Shareshian Jun 22 '20 at 15:31

Here are infinitely many examples showing that the answer to question 1 is negative. Take $n \equiv 1 \bmod 4$ with $n>5$ and let $\chi$ be the character of the symmetric group $S_n$ associated to the partition $(n-2,2)$. Then, as is well-known, for each $w \in S_n$, $\chi(w)$ is obtained by subtracting the number of fixed points of $w$ from the number of $2$-sets fixed (setwise) by $w$. In particular, $\chi(1)={{n} \choose {2}}-n=\frac{n(n-3)}{2}$, and if $w$ has cycle type $(n-4,4)$ then $\chi(w)=0$. Now $w$ generates its own centralizer, which thus has order $4(n-4)$, which is manifestly coprime with $\frac{n(n-3)}{2}$ under the given conditions.

  • $\begingroup$ @JohnShareshian Do you expect the existence of an example with $G$ an alternating group (or more generally a simple group)? $\endgroup$ – Sebastien Palcoux Jun 22 '20 at 17:22
  • $\begingroup$ I think that when $n=17$, you can use the same $\chi$ as in the answer (restricted to $A_n)$ and $w$ of shape $(9,4,4)$ to get an example in a simple group. There are likely infinitely many examples like this. $\endgroup$ – John Shareshian Jun 22 '20 at 17:33

A partial answer to Question 2: the following is a theorem of Burnside (see e.g. Isaacs, Theorem 3.8).

Theorem. Let $\chi$ be an irreducible character, let $K$ be a conjugacy class of $G$, and let $g\in K$. Suppose that $\gcd(\chi(1),\#K)=1$. Then either $\chi(g)=0$ or the irreducible representation with character $\chi$ sends $g$ to a scalar (i.e. $g \in {\rm Z}(\chi)$).

The proof uses the integrality property of central characters that John Murray mentions.

Since $\#K=\#G/\#{\rm C}_G(g)$, the hypothesis translates to the assumption that for every prime $p$ that divides $\chi(1)$, the centraliser ${\rm C}_G(g)$ contain a Sylow $p$-subgroup of $G$, so it is a stronger variant of the hypothesis in your Question 2. If for all $p|\chi(1)$ the Sylow $p$-subgroups of $G$ have order $p$, then the hypothesis of Burnside and of your question coincide.


Here is another beautiful little lemma that relates centralizers to zeros of characters. This can be easily read off the character table. It is due to Feit and Thompson and appeared in their Odd-Order Theorem paper.

Lemma Let $N \unlhd G$ and $\chi \in Irr(G)$ such that $N \not \subseteq ker(\chi)$. If $g \in G$ with $N \cap C_G(g)=1$, then $\chi(g)=0$.

Proof $Irr(G)$ can be split up into two disjoint subsets: $Irr(G)=S \mathop{\dot{\cup}} T$, where $S=\{\chi \in Irr(G): N \subseteq ker(\chi)\}$ and $T=\{\chi \in Irr(G): N \not\subseteq ker(\chi)\}$. Observe that $S$ can be identified with $Irr(G/N)$. If $g \in G$ with $N \cap C_G(g)=1$, then $C_G(g)$ embeds isomorphically into $C_{\overline{G}}(\overline{g})$ (where $\overline{.}$ denotes modding out by $N$), whence $|C_G(g)| \leq |C_{\overline{G}}(\overline{g})|$. By applying the Second Orthogonality Relation twice we get $$|C_{\overline{G}}(\overline{g})|=\sum_{{\chi \in S}}|\chi(\overline{g})|^2=\sum_{{\chi \in S}}|\chi(g)|^2 \geq |C_G(g)|=\sum_{{\chi \in Irr(G)}}|\chi(g)|^2=\sum_{{\chi \in S}}|\chi(g)|^2 + \sum_{{\chi \in T}}|\chi(g)|^2$$ yielding $\chi(g)=0$ whenever $N \not\subseteq ker(\chi)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.