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.
 A: 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.
A: 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. 
A: 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.
