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Is there a function $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for each finite supersolvable group $G$, and a Sylow subgroup $S \leq G$ we have $d(S) \leq f(d(G))$?

Here $d(H)$ denotes the minimal cardinality of a generating set of a group $H$. It is enough to consider the case of $G$ having a trivial Frattini subgroup.

In the language of profinite groups, an equivalent reformulation would be:

Let $F$ be a free finitely generated nonabelian prosupersolvable group. Is there a bound on the number of generators of the Sylow subgroups of $F$? (the sylow subgroups must be finitely generated by a theorem of Oltikar and Ribes).

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  • $\begingroup$ This may be interesting to know that the converse is true; that is, $d(G)\leq max_p(d(S_p))+1$ for any finite group $G$, where $S_p$ denotes a Sylow $p$-subgroup of $G$. See [R. M. Guralnick, On the number of generators of a finite group, Arch. Math. (Basel) 53 (1989), 521–523] and [A. Lucchini, A bound on the number of generators of a finite group, Arch. Math. (Basel) 53 (1989), 313–317.] $\endgroup$ – Alireza Abdollahi Jun 18 '14 at 6:54
  • $\begingroup$ Yes I knew this, and this was in fact the reason I asked the question. I wanted to give a bound on the number of generators of subgroups having the same Sylow subgroups for certain primes. $\endgroup$ – Pablo Jun 20 '14 at 5:33
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I don't think so. Let $p$ and $q$ be primes with $q|p-1$. Then there are $q$ inequivalent $1$-dimensional modules for $C_q$ over ${\mathbb F}_p$, If we take the semidirect product of the direct sum of these modules by $C_q$, then we get a supersolvable group $G$ of order $p^qq$ with $d(G)=2$, with an elementary abelian Sylow $p$-subgroup $P$ of order $p^q$, so $d(P)=q$. In terms of presentations, suppose that $\langle x,y \mid x^p=y^q=1, x^y=x^a \rangle$ is a presentation of a nonabelian group of order $pq$. Then we can take

$$G = \langle x_i,y \mid y^q=1, x_i^p=1, [x_i,x_j]=1, x_i^y = x_i^{a^i} \rangle$$

where $1 \le i,j \le q$. Then $G = \langle x_1x_2 \cdots x_q, y \rangle$, so $d(G)=2$.

PS: in fact there is no need to assume that $q$ is prime.

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