Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
5  
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
add comment

2 Answers

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.

share|improve this answer
1  
How did Berkovich miss this? –  Gerry Myerson Oct 15 '13 at 22:07
3  
@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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.