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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?

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  • $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Commented Feb 15, 2019 at 17:34

1 Answer 1

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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.

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