# Bounding the derived length of a solvable group given the degrees of the irreducible monomial characters

Much is known about the derived length of a solvable group given the degrees and cardinality of the set of degrees of the irreducible characters. Martin Isaacs and Donald Passman pretty much started this area of study in 1960.

Say you are only given the set of degrees of the monomial irreducible characters of a solvable group. Then how much is known about its derived length? Let's denote this set by m.c.d(G). It is not too difficult to show that if |m.c.d(G)|=1 then G must be abelian. Is it known that if |m.c.d(G)|=2 then G is metabelian for example? Or perhaps something similar?

• Do you define $\mathrm{mcd}(G)$ as a set (thus forgetting multiplicities) on purpose? – YCor Oct 28 '18 at 13:14
• Yes, we count without multiplicity. – Joakim Færgeman Oct 28 '18 at 13:28
• OK thanks. Have you done some tests in small groups? a computer should be able to compute mcd, say, for all groups of order 128 (relying on groupprops.subwiki.org/wiki/Groups_of_order_128, there are 2328, but only 14 being non-metabelian — groups of order dividing 64 are metabelian). – YCor Oct 28 '18 at 13:38
• Well, a group of order 128 is nilpotent and hence an M-group. So every irreducible character is monomial, and so its characters only have two different degrees - and M. Isaacs has shown that this implies that G is metabelian. I have not tried to do tests for small groups, though. – Joakim Færgeman Oct 28 '18 at 13:56
• sorry, I'm confused now: you seem to claim that every group of order 128 is metabelian, which is not true, so I probably misunderstand your comment. (There exist finite nilpotent groups whose set of degrees of irreducible characters is arbitrary large.) – YCor Oct 28 '18 at 15:11

I did a quick check in GAP, and unfortunately, it is not true that having $$2$$ monomial character degrees implies being metabelian.

A counter example is given by $$G = SL_2(3)$$ of order $$24$$, which has character degree $$\{1,2,3\}$$, but no monomial characters of degree $$2$$, so $$m.c.d(G) = \{1,3\}$$.

The derived length of $$G$$ is $$3$$, and unfortunately, the derived subgroup is of order $$8$$, so it is nilpotent, and hence the statement also fails even under the added assumption of a nilpotent derived subgroup.

• This really surprises me since when I tried to do a proof by induction, a group with |mcd(G)|=2 and a nilpotent derived subgroup has some truly weird and surprising properties. But thank you for the check - that certainly saved me some time! I think it is possible to make quite a good bound on the derived length, though. – Joakim Færgeman Oct 28 '18 at 19:22
• Quite possibly, yes. I don't have time right now, but the GAP code I used to check can easily be re-purposed to check various conjectured bounds like these. – Tobias Kildetoft Oct 28 '18 at 19:26
• @JoakimFærgeman I ran the code for a while, and up to order $300$ all solvable groups satisfy that their derived length is bounded by $|m.c.d| + 1$. I will check some more orders now, but for one of the orders between 300 and 400 my computer ran out of memory, so I need to split it up. – Tobias Kildetoft Oct 30 '18 at 7:58
• @JoakimFærgeman Ran it some more now. It still holds up to order $600$, except possibly for orders $384$ and $576$ which turned out to be too memory intensive to test naively like this. – Tobias Kildetoft Oct 30 '18 at 15:48
• This is very interesting - thank you for the run. If this is true in general, I do not suppose that it is any easier to proof than the Taketa Inequality. I think it would be natural (and doable) to try to get a bound for the derived length for groups of odd order first. – Joakim Færgeman Oct 30 '18 at 17:51