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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).

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  • $\begingroup$ 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. $\endgroup$ – Jeremy Rickard Nov 20 '18 at 13:49
  • $\begingroup$ 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 $n=22,23,24$. $\endgroup$ – John Murray Nov 20 '18 at 14:23
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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.

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    $\begingroup$ 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 $G\backslash\Phi(G)$. $\endgroup$ – John Murray Nov 21 '18 at 15:59

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