# character degree and solvability

There is an unsolved problem in Berkovich's book "Characters of Finite Groups Part 2" I state here:

Is $G$ solvable if $\chi(1)^2$ divides $|G|$ for all $\chi \in {\rm Irr}(G)$?

Can any one tell me some latest progresses for this? Maybe you can tell me some latest research papers. Thank you.

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Cross-posted from MSE: math.stackexchange.com/questions/517384/… You could look at the edits to your question there to see how to format it properly. –  Tobias Kildetoft Oct 7 '13 at 12:16

No!$~~~$If $S$ is any finite group, let $G = S \times A$, where $A$ is abelian and $|A| = |S|$. Then the degree of each irreducible character of $G$ is the degree of an irreducible character of $S$, so it divides $|S|$. Its square, therefore, divides $|G|$. If $S$ is not solvable, then, of course, $G$ is nonsolvable too.

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How did Berkovich miss this? –  Gerry Myerson Oct 15 '13 at 22:07
@GerryMyerson Indeed. And how did the rest of us do the same? –  Tobias Kildetoft Oct 16 '13 at 7:28
Marty, great counterexample! –  Nicky Hekster Oct 22 '13 at 9:36

In the nice counterexample that Marty gave, the fact that $\chi(1)$ divides $[G:Z(G)]$ when $\chi$ is a complex irreducible character of the finite group $G$ is exploited. Here is a similar example, where the group $G$ has $Z(G)= 1.$ It exploits a theorem of Ito which asserts that $\chi(1)$ divides $[G:A]$ for each Abelian normal subgroup $A$ of $G.$ Take $H = A_{5} \cong {\rm SL}(2,4) \cong {\rm PSL}(2,5).$ Then $H$ acts non-trivially on an elementary Abelian $2$-group $U$ of order $16$ and also acts non-trivially on an elementary Abelian $3$-group $V$ of order 729. Also, $H$ acts non-trivially on an elementary Abelian $5$-group $W$ of order $125$. Let $G$ be the semi-direct product $( U \times V \times W).H,$ with $H$ acting faithfully on each factor. Then $|G|$ is divisible by $3600$ and each complex irreducible character of $G$ has degree dividing $60$. In this case $Z(G)= 1.$ Hence strengthening Berkovich's question to the case that $\chi(1)^{2}$ divides $[G:Z(G)]$ for each complex irreducible character $\chi$ still yields a negative answer. Perhaps one could ask whether $G$ is solvable if $\chi(1)^{2}$ divide $[G:A]$ for each Abelian normal subgroup $A$ of $G$ and each irreducible complex character $\chi$ of $G.$ However, that is an extremely strong hypothesis, and even when $G$ is a $p$-group, it need not be satisfied.

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Steve Gagola and I have a paper where we address this very question. I am away from home and so I don't have the reference handy. It was in the Communications in Algebra in the late 90's. I think 1998.

Also, the example Marty gave was in that paper. Marty and I found that example while I was still working on my PhD.

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Perhaps Gagola, Stephen M., Jr.; Lewis, Mark L., A character-theoretic condition characterizing nilpotent groups, Comm. Algebra 27 (1999), no. 3, 1053–1056, MR1669104 (2000a:20011). –  Gerry Myerson Jun 17 '14 at 4:03
Yes. That is the one. –  Mark Lewis Jun 19 '14 at 18:18
In his review in Zentralblatt Zbl 0929.20010 of our paper, Berkovich conjectured that if χ(1)^2 divides |G|>1 for all χ∈Irr(G), then F(G)>{1}, where F(G) is the Fitting subgroup of G. If so, this result is a generalization of Theorem A, in view of Proposition 2.2. In the following paper, Gagola showed that this conjecture is true. Gagola, Stephen M., Jr.(1-KNTS) A character theoretic condition for F(G)>1. Comm. Algebra 33 (2005), no. 5, 1369–1382 –  Mark Lewis Jun 19 '14 at 18:23
We actually proved: A finite group G is nilpotent if and only if chi (1)^2 divides |G:ker (chi)| for all chi in Irr (G). –  Mark Lewis Jun 19 '14 at 18:38