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For a finite group $G$ we denote $d(G)$ the minimal size of a set of generators of $G$. We define $D(G) = \max( d(H) \mid H\leq G)$.

Let $S$ be a finite simple group. Are there `good' bounds on $D(S)$ in terms of the size of $S$?

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    $\begingroup$ Every finite group $G$ embeds into $A_n$ for some $n>4$. This gives some (not very good) bound for $D(S)$. $\endgroup$
    – user6976
    Jun 23, 2012 at 13:08
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    $\begingroup$ $\log_2 |S|$ is an easy upper bound. I would be surprised if the large elementary abelian groups of rank about $n/2$ inside the alternating groups $A_n$ didn't provide the asymptotic maximum. $\endgroup$ Jun 23, 2012 at 13:10
  • $\begingroup$ @Duglas: You are probably right that the Abelian subgroups have maximal rank, but perhaps one should consider groups of Lie type instead of $A_n$. $\endgroup$
    – user6976
    Jun 23, 2012 at 13:56
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    $\begingroup$ No it seems that $A_n$ is better asymptotically ($n!$ vs $q^{n^2}$). $\endgroup$
    – user6976
    Jun 23, 2012 at 14:24
  • $\begingroup$ by mistake I rolled back, and can't find how to undo. HELP!!! $\endgroup$ Nov 25, 2012 at 10:12

3 Answers 3

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By a Theorem of R. Guralnick and A. Lucchini (see MR1015993 and MR 1023965),(which does require the classification of finite simple groups) the minimum number of generators for a finite group $G$ can exceed by at most one the maximum (over all its Sylow subgroups $P$) of the minimum number of generators of $P$. It follows that the value of $D(G)$ is between $d(H)$ and $d(H)+1$ for some $p$-subgroup $H$ of $G.$ Thus for a finite simple group $S,$ the question does essentially come down to bounding the minimum number of generators of subgroups of $S$ of prime power order, as was suggested might be the case in some comments. The sectional $p$-rank of a finite group $G$ is the maximum over all section of $G$ which are $p$-groups, of the minimum number of generators of that section (a section of $G$ is a group of the form $H/K$ where $H$ is a subgroup of $G$ and $K \lhd H ).$ Hence if we define the sectional rank of $G$ to be the maximum of the minimum number of generators of an Abelian section of $G$, and denote it by $ar(G),$ then we see that for any finite group $G$, simple or not, we have $ ar(G) \leq D(G) \leq ar(G)+1.$

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  • $\begingroup$ For finite simple groups sectional rank and the ordinary rank (the biggest rank of an abelian subgroup) almost coincide as well? $\endgroup$
    – user6976
    Jun 24, 2012 at 5:08
  • $\begingroup$ Thank you very much for the answer. Is there something known about $ar(S)$ for finite simple groups? $\endgroup$ Jun 24, 2012 at 6:28
  • $\begingroup$ @Mark: I am not sure. $\endgroup$ Jun 24, 2012 at 10:05
  • $\begingroup$ Note that the minimum number of generators of a finite Abelian group is the maximum of the same quantity over its Sylow $p$-subgroups. The structure of Sylow $p$-subgroups of a finite simple group can be analyzed in any particular case. $\endgroup$ Jun 24, 2012 at 10:11
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    $\begingroup$ @GeoffRobinson no, you misunderstood my comment. I your example $D(G)$ is equal to $D(H)$ where $H$ is a $p$-Sylow in $G$. (The point of your example is that $D(H)$ is not equal to the rank of $H$ itself, which is smaller.) Here's an equivalent reformulation of my comment. Let $D(G)$ be the sup of ranks of all subgroups of $G$, and $D'(G)$ the sup of ranks of all subgroups of prime-power order of $G$. Is the inequality $D'(G)\le D(G)$ always an equality? (or "close" to an equality: bounded error, few exceptions, etc) $\endgroup$
    – YCor
    Apr 23, 2023 at 10:52
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Every finite simple group can be generated by two elements. Except in the case of prime order, one of the elements can have order 2. See here for example.

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    $\begingroup$ This does not answer the question. $\endgroup$ Jun 23, 2012 at 12:38
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    $\begingroup$ @Andreas: Serves me right for not reading it properly. Thanks. $\endgroup$ Jun 24, 2012 at 2:53
  • $\begingroup$ but interesting nonetheless! $\endgroup$
    – Will Chen
    Feb 10, 2015 at 23:04
  • $\begingroup$ The link to springerlink.com is broken. I'm also unable to find any snapshot saved on the Wayback Machine. $\endgroup$ Apr 17, 2023 at 10:06
  • $\begingroup$ mathoverflow.net/questions/59213/… seems to be relevant. $\endgroup$ Apr 17, 2023 at 12:20
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Nobody seems to have mentioned the work of Burness, Liebeck and Shalev yet:

Burness, Timothy C.; Liebeck, Martin W.; Shalev, Aner, Generation and random generation: from simple groups to maximal subgroups., Adv. Math. 248, 59–95 (2013). Zbl 1292.20013. PDF

They prove that if $S$ is a non-abelian finite simple group and $H$ is a maximal subgroup of $S$ then $d(H)\leq4$. Furthermore, there are infinitely many examples that attain this bound.

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