Does $\chi(1)^2=|G:Z(G)|$ for irreducible character of a finite group $G$ imply $G$ is solvable?

Question

In "Character Theory of Finite Groups" I.M. Isaacs mention the following conjecture:

It is only possible in a solvable group $G$ to have $\chi(1)^2=|G:Z(G)|$ with $\chi \in$ Irr$(G)$.

Is this problem still open? I tried to search for attempts to solve it but didn't find anything.

Frieder Ladisch · Accepted Answer · 2020-03-18 02:28:27Z

13

Such groups are solvable. This has been solved by Howlett and Isaacs himself, in

Howlett, Robert B.; Isaacs, I. Martin, On groups of central type, Math. Z. 179, 555-569 (1982). MR652860 ZBL0511.20002.

The proof uses the classification of finite simple groups.

answered Mar 18, 2020 at 1:34

Frieder Ladisch

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$\begingroup$ It may be trivial but could you clarify why groups with the OP condition are of central type? $\endgroup$
– user6976
Commented Mar 18, 2020 at 2:07
$\begingroup$ It is clearer in the MathSci review of the paper. $\endgroup$
– user6976
Commented Mar 18, 2020 at 2:18
$\begingroup$ @MarkSapir This is one of several equivalent definitions of "of central type" (in this context), see the first two sentences in the cited paper. That these are equivalent is shown in Cor. 2.30 of Isaacs's book, from where the citation is. $\endgroup$
– Frieder Ladisch
Commented Mar 18, 2020 at 2:19
$\begingroup$ Thanks! I understood it after I read the MathSci review. $\endgroup$
– user6976
Commented Mar 18, 2020 at 2:22
$\begingroup$ Added link to MathSci review. The link to Zentralblatt was suggested automatically to me when I used the citation feature. $\endgroup$
– Frieder Ladisch
Commented Mar 18, 2020 at 2:30

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