The zero entries in the character table of a finite group

Question

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)| $?

Answer 1

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.

Answer 2

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.

Answer 3

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 $N \not\subseteq ker(\chi)$.