Questions about a finite solvable group

These questions are by Moshe Newman

Let $$G$$ be a finite solvable group of derived length $$d$$, with the property that every proper subgroup and every proper quotient of $$G$$ has derived length less than $$d$$.

1) Can one give an upper bound on the number of different prime divisors of $$|G|$$, the order of $$G$$, a bound that depends only on $$d$$?

2) Can one give an upper bound on the number of elements required to generate $$G$$, a bound that depends only on $$d$$?

For $$d=2$$ the answer to both questions is yes, and the bound is $$2$$ in both questions. (See Miller and Moreno, for example.) What happens in general? Comment : one cannot bound the composition length, even for the case d = 2. Thus one can only hope to bound the number of distinct prime divisors of $$|G|$$, and not the total number of prime divisors counting multiplicity. Comment : assuming the answer to 2) is yes, it is the case that the bound must grow with $$d$$. Is this true for 1) as well?

• I think you can construct examples for a given $d$ that are divisible by $d$ distinct primes, so the answer to the very final question is yes. Construct a sequence of groups $S_1,S_2,\ldots$ with $S_1 = C_2$, where each $S_k$ is divisible by the first $k$ primes. For $k>1$, let $S_k$ be the semidirect product of a faithful irreducible ${\mathbb F}_{p_k}S_{k-1}$-module with $S_{k-1}$, where $p_k$ is the $k$th prime. Since each $S_k$ has a unique minimal normal subgroup, such a module always exists. – Derek Holt Feb 15 '19 at 17:34

1 Answer

This partial solution by Stuart Margolis:

The answer to question $$2$$ is definitely yes, the number of elements required to generate $$G$$ can't be more than $$D=2^{d-1}$$. Suppose that $$G$$ requires more than $$D$$ generators, then every $$D$$-tuple of elements of $$G$$ must belong to a proper subgroup, which therefor must be solvable of derived length at most $$d-1$$. So every $$D$$-tuple of elements satisfies the identity (in $$D$$ variables) that defines being solvable of derived length at most $$d-1$$. So $$G$$ itself must be solvable of derived length at most $$d-1$$.

This is just a special example of the fact that if a variety (in the Hanna Neumann sense) is defined by an $$n$$-variable law then: a group which is not in the variety, but all its subgroups do belong to the variety, must be generated by at most $$n$$ elements.