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Can one find a function $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for every finite supersolvable group $G$ we have: $d(G') \leq f(d(G))$?

Here $d(K)$ is the cardinality of a minimal set of generators for $K$, and $K'$ is the commutator subgroup of $K$.

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Strictly, I think you mean $d(G)$ is the minimal cardinality of a generating set of $G$. There are usually minimal (ie irredundant) generating sets of $G$ of different sizes- unless you are using some special property of supersolvable groups which I am not aware of. –  Geoff Robinson May 31 at 15:04
    
Yes, definitely. –  Pablo May 31 at 22:18

2 Answers 2

up vote 3 down vote accepted

By a method similar to Jeremy Rouse's comment on my original (incorrect) answer, I think the answer is no. A $p$-group is certainly supersolvable. If we take $G = C_{p} \wr C_{p^{n}},$ then $d(G) =2,$ $d(G^{\prime}) = p^{n}-1$ - in fact, the case $n = 1$ already seems to provide a counterexample to the question as asked, since $p$ is an arbitrary prime, but this shows that there is no bound in general, even if we allow a function which depends on the prime divisors as well as $d(G).$

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I just convinced myself this can't be true in an arbitrary solvable group. Fix an arbitrary $n$ and choose an $n \times n$ matrix $M$ with entries in $\mathbb{F}_{2}$ that has an irreducible characteristic polynomial and let $G = \mathbb{F}_{2}^{n} \rtimes \langle M \rangle$. Isn't it the case that $d(G) = 2$ and $d(G') = n$? Of course, this group isn't supersolvable. –  Jeremy Rouse May 31 at 1:31
    
Oops, yes, I remembered Segal's theorem incorrectly. He proves that every element of $G^{\prime}$ is a product of at most $72d^{2}+46d$ commutators, but he does not get a bound for the number of generators of $G^{\prime}.$ –  Geoff Robinson May 31 at 8:05
    
@Jeremy Rouse: You are right I had completely misremembered what Segal had proved. Your example was quite close to a counterexample to the original question actually. –  Geoff Robinson May 31 at 8:17

Let $p$ be a prime number, $2 \leq d \in \mathbb{N}$. Take $F$ to be the free pro-$p$ group on $d$ generators. $F'$ is an infinite index normal subgroup of $F$ so it must be infinitely generated. This means that there is some finite image (which is supersolvable) in which the the number of generators is arbitrarily high.

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