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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.

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    $\begingroup$ This rank was introduced by Malcev and is called special rank. Namely, The special rank of a group $G$ is the minimal $d$ such that every finitely generated subgroup of $G$ can be generated by $d$ elements. $\endgroup$
    – Anton Klyachko
    Dec 6, 2014 at 20:04

2 Answers 2

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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 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

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    $\begingroup$ Excellent! For a complete answer you need to argue why treating the case $n=2$ (as you did) is sufficient (it is possible that there are relations involving more than 2 elements). I think that such an argument has been given by @HJRW in the comments to his answer. $\endgroup$
    – Pablo
    Jan 28, 2015 at 10:14
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    $\begingroup$ Great! I didn't know that $SL_2(F_p)\in\mathcal{F}_2$ for all $p$. $\endgroup$
    – HJRW
    Jan 28, 2015 at 10:18
  • $\begingroup$ What happens if we fix a prime $p$, and consider only finite $p$-groups and not all finite groups? This was the last comment in my question. $\endgroup$
    – Pablo
    Jan 28, 2015 at 10:56
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    $\begingroup$ @Pablo Possibly you can adapt a proof that $F_2$ is residually $p$. Replace every occurence of $\mathbf{F}_p$ above with $\mathbf{Z}/p^n\mathbf{Z}$, now with $p$ fixed, and replace $A$ and $B$ respectively with $(1,p,0,1)$ and $(1,0,p,1)$. Then for $n$ large enough depending on your word the group $\langle A,B\rangle$ might suffice, but this requires better knowledge of $\text{SL}_2(\mathbf{Z}/p^n\mathbf{Z})$ than I claim to have. Maybe somebody else can comment. $\endgroup$
    – Sean Eberhard
    Jan 28, 2015 at 12:05
  • $\begingroup$ Thinking about it again after lunch, I think it's much more likely that if $G\in\mathcal{F}_d$ is nilpotent then its nilpotency class must be bounded in terms of $d$, which would of course imply that there is such a word. $\endgroup$
    – Sean Eberhard
    Jan 28, 2015 at 13:28
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As I said in a comment on your other questions, there is no such word. Indeed, for any $w\in F_n$, there is $u\in F_{n-1}$ so that $w$ survives under the retraction $F_n\to F_{n-1}$ that sends the last generator to $u$.

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  • $\begingroup$ And why there are no such words $w(x,y)\in F_2$? $\endgroup$
    – Anton Klyachko
    Dec 4, 2014 at 14:17
  • $\begingroup$ @AntonKlyachko - good point! Because $F_2$ is residually finite. $\endgroup$
    – HJRW
    Dec 4, 2014 at 14:37
  • $\begingroup$ @HJRW Have you noticed that I required every subgroup of $G$ to be generated by at most $d$ elements and not just $G$ itself? I just fail to see why does your answer solve my question... $\endgroup$
    – Pablo
    Dec 4, 2014 at 14:42
  • $\begingroup$ @Pablo - I'm sorry, you're right - I hadn't noticed this requirement. I guess my answer shows that you can take $n=d=2$. $\endgroup$
    – HJRW
    Dec 4, 2014 at 14:47
  • $\begingroup$ @HJRW I don't think I see how, but if this is true I find it very interesting... $\endgroup$
    – Pablo
    Dec 4, 2014 at 14:55

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