# Divisors of the regular character of a finite group

Recall that the regular character $$\rho=\hspace{-.2cm}\sum\limits_{\chi\in\operatorname{Irr}(G)}\hspace{-.2cm}\chi(1)\chi$$ of a finite group $$G$$ takes values $$\rho(g)= \left\{\begin{array}{cl} |G|,&\quad\mbox{if g=1.}\\ 0,&\quad\mbox{if g\in G\backslash\{1\}.} \end{array} \right.$$ Given $$\alpha\in\operatorname{Irr}(G)$$, we shall say that $$\alpha$$ divides $$\rho$$ if $$\alpha\beta=\rho$$, for some generalized character $$\beta$$.

Now $$\alpha\rho=\alpha(1)\rho$$. So linear characters of $$G$$ divide $$\rho$$. Notice that if $$\alpha$$ is faithful, then there is an integer polynomial $$p$$ such that $$\alpha p(\alpha)=n_\alpha\rho$$, for some positive integer $$n_\alpha$$. But I don't know anything about $$n_\alpha$$. I have found only a small number of cases of $$G$$ and $$\alpha$$ where $$\alpha$$ does not divide $$\rho$$ (see below). My question is a small variant on https://math.stackexchange.com/questions/2933550/tensor-complement-of-representations-of-finite-groups:

Question: Does there exist a $$p$$-group $$G$$ and $$\alpha\in\operatorname{Irr}(G)$$, such that $$\alpha$$ does not divide $$\rho$$?

Note that in my original question solvable group was in place of $$p$$-group. As Jeremy Rickard has pointed out, there are 3 groups of order $$72$$ with irreducible characters which do not divide $$\rho$$ (SmallGroups(72,n), for $$n=22,23,24$$, in GAP notation).

Non-solvable example: Let $$\alpha$$ be one of the two degree 4 irreducible characters of $$\operatorname{SL}(2,5)$$. Then $$\alpha$$ has a single conjugacy class of zeros $$4a$$. Let $$\beta$$ be a class function such that $$\alpha\beta=\rho$$. Then $$\beta(1a)=30$$, $$\beta(4a)$$ is odd and $$\beta$$ vanishes on all other classes. Let $$\psi\in\operatorname{Irr}(\operatorname{SL}(2,5))$$, with $$\psi(1)=2$$. Then $$\psi(4a)=0$$. So $$\langle\beta,\psi\rangle=\frac{1}{2}$$, showing that $$\beta$$ is not a generalized character.

A similar example is provided by a degree $$9$$ irreducible character of the group $$3.A_6.2_2$$ (the second degree 2 extension of the triple cover of the alternating group $$A_6$$, in Atlas notation). The groups $$\operatorname{SL}(2,5)$$ and $$3.A_6.2_2$$ are in the small list of non-solvable groups with an irreducible character which vanishes on only one conjugacy class. See S.~Madanha, On a question of Dixon and Rahnamai Barghi, arXiv:1811.03972 [math.GR].

On a positive note, $$\alpha$$ divides $$\rho$$ in the following cases:

1. $$\alpha(1)=|G|_p$$, for some prime $$p$$ (take $$\beta=1_S^G$$, for $$S\in\operatorname{Syl}_p(G)$$).

2. $$\alpha$$ is totally ramified with respect to an irreducible character of $$Z(G)$$.

3. $$\alpha$$ is induced from a linear character of a normal subgroup of $$G$$ with cyclic quotient (R. Gow).

4. $$G$$ is a $$2$$-group with $$|G|\leq256$$ or a $$3$$-group with $$|G|\leq729$$ (GAP). In fact for all such $$G$$ and $$\alpha$$, it seems that there is a character $$\beta$$ with $$\alpha\beta=\rho$$.

5. $$G=A_n$$ or $$S_n$$, for $$n\leq15$$ (GAP).

6. $$G=\operatorname{SL}(2,3),\operatorname{GL}(2,3)$$ or the binary octahedral group (fake $$\operatorname{GL}(2,3)$$').

7. $$G$$ is a small finite simple group (GAP).

• You might be interested in this question on math.stackexchange: math.stackexchange.com/questions/2933550/… When I answered that question I found solvable examples of order 72 and 120 by running through GAP's SmallGroups. – Jeremy Rickard Nov 20 '18 at 13:49
• Thanks very much Jeremy. “Tensor complement of representations of finite groups" asks almost the same question, except for their requirement that $\beta$ be a character. The solvable examples you mention are SmallGroups(72,n), for $n=22,23,24$. – John Murray Nov 20 '18 at 14:23

I am not sure if it helps, but I think in a minimal $$p$$-group $$P$$ which has an irreducible character $$\alpha$$ not being a divisor of $$\rho,$$ we may suppose that $$\alpha$$ vanishes identically outside $$\Phi(P).$$

For suppose there is $$x \in P \backslash \Phi(P)$$ with $$\alpha(x) \neq 0.$$ We may choose a maximal subgroup $$M$$ of $$P$$ with $$x \not \in M.$$ Then $${\rm Res}^{P}_{M}(\alpha)$$ is irreducible ( for otherwise $$\alpha$$ is induced from an irreducible character of $$M$$ and vanishes identically outside $$M$$ (so, in particular, $$\alpha(x) = 0,$$ contrary to assumption)).

By the minimal choice of $$P,$$ there is a generalized character $$\gamma$$ of $$M$$ with $${\rm Res}^{P}_{M}(\alpha) \gamma = \rho_{M},$$ the regular character of $$M$$.

Now for each $$y \in P \backslash M,$$ both $${\rm Res}^{P}_{M}(\alpha)$$ and $$\rho_{M}$$ are $$y$$-stable, so we also have $${\rm Res}^{P}_{M}(\alpha) \gamma^{y} = \rho_{M}.$$

Now it follows that $$\alpha {\rm Ind}_{M}^{P}(\gamma) = \rho,$$ since this generalized character certainly vanishes identically outside $$M,$$ and agrees with $$p\rho_{M}$$ on $$M.$$ Hence $$\alpha$$ is a divisor of $$\rho,$$ contrary to assumption.

• Thanks Geoff, that argument is useful even if $G$ is not a $p$-group. For Jeremy's SmallGroup(72,n), with $n=22,23,24$, we have $G=E_9:S$, where $S=D_8$ or $Q_8$, and $\Phi(G)=E_9:C_2$. The bad' irreducible characters have degree $4$. In each case their zero set is $G\backslash\Phi(G)$. – John Murray Nov 21 '18 at 15:59