Let $d \geq 2$ be an integer, and let $\mathcal{F}\_d$ be the family of finite groups such that $G \in \mathcal{F}\_d$ if and only if every subgroup of $G$ can be generated by at most $d$ elements. 


**Is there a finite nontrivial word $w = w(x\_1, \dots, x\_n)$ which is trivial on $\mathcal{F}\_d$?** 


That is, can it be that for any choice of $G \in \mathcal{F}\_d$ and $y\_1, \dots, y\_n \in G$ we have $w(y\_1, \dots, y\_n) = 1$.


Alternatively: Let $F$ be a free group of rank $\aleph\_0$. Is it possible that the intersection of all finite index subgroups $N \lhd F$ with $F/N \in \mathcal{F}\_d$ is nontrivial? 


I am also interested in the analogous question for $p$-groups.



Is the free abstract group residually of rank d > 2?
No, there is not such a word, by the following two facts.


1. The two elements
$$A=\left(\begin{array}{cc}1&2\\0&1\end{array}\right),\quad B=\left(\begin{array}{cc}1&0\\2&1\end{array}\right)$$ of $\text{SL}\_2(\mathbf{F}\_p)$ satisfy no nontrivial relation of length less than $c\log p$.
2. Every subgroup of $\text{SL}\_2(\mathbf{F}\_p)$ is generated by at most 2 elements.


For 1, there is a well known [ping-pong argument](http://en.wikipedia.org/wiki/Ping-pong_lemma#Special_linear_group_example) which proves that $A$ and $B$ generate a free subgroup of $\text{SL}\_2(\mathbf{Z})$. The argument can be reproduced in $\text{SL}\_2(\mathbf{F}\_p)$ for words which are not too long.


For 2, the subgroups of $\text{SL}\_2$ are well understood. Refer for instance to Theorem 6.17 in Suzuki's book:


Suzuki, Michio. Group theory. I. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 247. Springer-Verlag, Berlin-New York, 1982. xiv+434 pp. ISBN: 3-540-10915-3 MR0648772 


