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Definition: The subgroup rank of a finite group $G$ is the minimal natural number $n$ such that every subgroup of $G$ can be generated by $n$ elements (or fewer).

This invariant has been studied extensively for various families of groups. I am interested in the family of finite simple groups and I have been unable to find and relevant information in the literature.

Question 1: Are there only finitely-many finite simple, non-abelian groups $G$ of a given subgroup rank $n$?

Some relatively straight-forward comments and reductions:

It is not too difficult to show that there are only finitely-many alternating groups of subgroup rank at most $n$ (by explicitly constructing elementary-abelian subgroups of a certain subgroup rank). There are also only finitely-many sporadic groups, according to the classification. These observations reduce the above question to finite simple groups of Lie-type.

Question 2: Are there only finitely-many finite simple groups $G$ of Lie-type with given subgroup rank $n$?

It is again not too difficult to show that the "field rank" of $G$ is bounded from above by a function of $n$ (by looking at the natural homomorphism from the field to the root subgroups). It is also possible to show that the Lie-rank of $G$ is bounded from above by a function of $n$. These observations further reduce Question 2 to bounding the defining characteristic of the simple group of Lie-type by some function that depends only on the subgroup rank $n$. Unfortunately, I do not have any good intuition to determine whether the latter statement is true or not.

I hope both questions have a positive answer because that would give us a nice property about the FSG. But I suspect we can prove the answers to be "no" by simply making some judicious choice for the Lie-type, field-rank, and Lie-rank and by then looking at the structure of the Sylow-subgroups of $G$, as the characteristic goes through the different primes.

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  • $\begingroup$ Here's my answer to another question about the notion of bounded rank finite simple group. Clearly bounded subgroup rank implies bounded rank. The converse is not true since $\mathrm{PSL}_2(\mathbf{F}_{p^n})$ contains a copy of $(\mathbf{Z}/p\mathbf{Z})^n$, and hence, for varying $n$ has unbounded subgroup rank. However $\mathrm{PSL}_2(\mathbf{F}_p)$ has bounded subgroup rank when the prime $p$ varies, by this answer. $\endgroup$ – YCor May 4 at 15:42
  • $\begingroup$ It might be that a family of bounded rank finite simple groups has bounded subgroup rank iff the associated fields have bounded degree over their prime subfield. $\endgroup$ – YCor May 4 at 15:42
  • $\begingroup$ Dear @Yves, thank you for the comments. We seem to have given the same proof for the implication "bounded subgroup rank implies bounded Lie-rank and bounded field-rank." Here, my "Lie-rank" is your "rank," and my "field-rank" is your "degree over the prime field." I chose to use slightly unorthodox terminology in my original post to avoid mixing up the various different invariants. Apologies for the confusion. Concerning your conjecture about the reverse statement: I wondered the same after the posts of Geoff and Derek. Interesting question! $\endgroup$ – user203598 May 4 at 16:57
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    $\begingroup$ @YCor: Your conjecture is true. Use the Guralnick-Lucchini theorem in Geoff's answer. Let $G$ be the $F_{r^a}$-points of a simple algebraic group $\bar G$ of Lie rank $n$, and let $P$ be a Sylow $p$-subgroup of $G$. If $p\ne r$, $P\le N_{\bar G}(T)$ for some maximal torus $T$ of $\bar G$, so the $p$-subgroup rank of $P$ is at most $n+\log_2|W|$, independent of $a$ and $p$. If $p=r$, then $|P|\le p^{Ca}$ where $C$ is the number of positive roots, so the $p$-subgroup rank is at most $Ca$. $\endgroup$ – Richard Lyons May 7 at 17:58
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By a result of R.Guralnick and A. Lucchini (obtained independently, not in joint work) the subgroup rank of a finite group $G$ is bounded above by $1+d$, where $d$ is the maximum over all $p$-subgroups $P$ of $G$ (and over all primes $p$) of the minimum number of generators of $P$. This requires the classification of finite simple groups.

It follows that the subgroup rank of any of the simple groups ${\rm PSL}(2,r)$ ($r >3$ prime) is at most $3$, since every $p$-subgroup of such a group is either cyclic (if $p$ is odd) or cyclic, dihedral, or Klein $4$ when $p = 2$.

Hence the answer to the first question is "no" in general.

It would appear that this argument does rely on the field of definition of the Lie type group being small, and the Lie rank being small.

Later edit: There may be a more elementary proof of the subgroup rank at most $3$ property of ${\rm PSL}(2,r)$ (ie without CFSG), by work of L.E. Dickson.

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    $\begingroup$ Dickson gave a complete description of all subgroups of ${\rm PSL}(2,q)$ - for $q$ prime, they are cyclic, dihedral, affine or $A_4$, $S_4$, $A_5$ - so it should be straightforward to justify the answer no without using CFSG. It looks as though they are all $2$-generated. $\endgroup$ – Derek Holt May 4 at 11:00
  • $\begingroup$ Yes, thanks Derek, I had already remembered Dickson's work and put a note about it at the end. I agree that all subgroups of PSL(2,r) appear to be $2$-generated, though I haven't checked every detail. $\endgroup$ – Geoff Robinson May 4 at 11:06
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    $\begingroup$ @Geoff: Thank you for the quick reply. That answers both questions in the negative (not just the first one). $\endgroup$ – user203598 May 4 at 12:49

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