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Let $G$ be a finite group. Assume that $\chi$ is a complex irreducible character of $G$ of degree $n\geq 2$, with the property that for each element $g\in G$ either $\chi(g)=0$ or $|\chi(g)|=n$.

  1. Does it necessarily follow that $\chi$ is imprimitive, i.e., induced from a character of a subgroup?

The only such examples I could think of are given by extraspecial groups.

  1. Is there a classification of all the finite groups with this property?

The questions are partly motivated by this post: Finite groups with a character having very few nonzero values?

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    $\begingroup$ There is a whole literature on finite groups of central type, which is the name for the characters you describe. $\endgroup$
    – Geoff Robinson
    Commented Aug 8, 2018 at 11:44
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    $\begingroup$ I believe that any finite group with such a faithful character is solvable, and I believe the result is due to Howlett and Isaacs. $\endgroup$
    – Geoff Robinson
    Commented Aug 8, 2018 at 11:54
  • $\begingroup$ Thanks! I’m sure it’s probably obvious, but is there a quick way to see that my condition is equivalent to saying that $G$ has a character of degree equal to the square root of the index of the center? This seems to be the usual way groups of central type are defined. $\endgroup$
    – Goro
    Commented Aug 8, 2018 at 12:10
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    $\begingroup$ Just compute the character inner product $\langle \chi, \chi \rangle = 1$, which means that $\sum_{g \in G} |\chi(g)|^{2} = |G|$. On the other hand,$|\chi(g)|^{2} = \chi(1)^{2}$ for $g \in Z(G),$ and $0$ otherwise, so we obtain $|Z(G)|\chi(1)^{2} = |G|$ and $\chi(1)^{2} = [G:Z(G)].$ On the other hand, if $G$ has a faithful irreducible character $\mu$ with $\mu(1)^{2} = [G:Z(G)],$ then $\mu$ must vanish identically outside $Z(G)$ ( and we have $|\mu(z)| = \mu(1)$ for $z \in Z(G)).$ $\endgroup$
    – Geoff Robinson
    Commented Aug 8, 2018 at 13:28
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    $\begingroup$ I won't write this as a (partial) answer, but here are a couple of general properties of groups of central type: If G is of central type, then so is every Sylow p-subgroup ( a Sylow p-subgroup $P$ might be Abelian, in which case $P \leq Z(G)$, and $P$ is a direct factor of $G$. On the other hand, if every Sylow p-subgroup of $P$ $G$ is of central type (for every choice of $p$) with $Z(P) \leq Z(G),$ then $G$ is of central type ( this needs Brauer'c characterization of characters). Also the faithful character $\chi$ is in a $p$-block of full defect for every $p$. $\endgroup$
    – Geoff Robinson
    Commented Aug 9, 2018 at 10:49

1 Answer 1

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As Geoff said, a group having such an irreducible character is called a group of central type, and Howlett and Isaacs have shown, using the classification of finite simple groups, that such groups are solvable: see
Howlett, Robert B.; Isaacs, I. Martin, On groups of central type, Math. Z. 179, 555-569 (1982). ZBL0511.20002.

Using this result, we can show that a character $\chi$ of central type is necessarily imprimitive. For suppose that $Z$ is the center of $\chi$, so that $\chi(1)^2 = \lvert G:Z \rvert $, and let $N$ be any normal subgroup of $G$ containing $Z$. Let $\tau$ be a constituent of $\chi_N$. By Clifford theory, $\chi$ is induced from the inertia subgroup of $\tau$ in $G$. So if $\tau$ is not invariant in $G$, we are done. Otherwise, we have $\chi_N = e\tau$ and $\tau$ vanishes outside $Z$. It follows that $e^2 = \lvert G:N \rvert$ and $\lvert G:N \rvert $ is a square. But by solvability of $G$, maximal normal subgroups have prime index.

Of course it would be nice to have an argument not depending on CFSG.
To the best of my knowledge, groups of central type are not classified. The paper by Howlett and Isaacs contains a non-nilpotent example. There are also nilpotent examples of arbitrarily large nilpotency class: Take the semidirect product of a cyclic group of order $p^{n+1}$ with the Sylow $p$-subgroup of its automorphism group.

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