When you browse the character tables of the small finite groups (for example [here](https://people.maths.bris.ac.uk/%7Ematyd/GroupNames/characters.html)), 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)| $?



The zero entries in the character table of a finite group
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.


