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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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Every finite group $G$ embeds into $A_n$ for some $n>4$. This gives some (not very good) bound for $D(S)$. – Mark Sapir Jun 23 '12 at 13:08
$\log_2 |S|$ 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. – Douglas Zare Jun 23 '12 at 13:10
@Duglas: You are probably right that the Abelian subgroups have maximal rank, but perhaps one should consider groups of Lie type instead of $A_n$. – Mark Sapir Jun 23 '12 at 13:56
No it seems that $A_n$ is better asymptotically ($n!$ vs $q^{n^2}$). – Mark Sapir Jun 23 '12 at 14:24
by mistake I rolled back, and can't find how to undo. HELP!!! – Lior Bary-Soroker Nov 25 '12 at 10:12
up vote 12 down vote accepted

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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For finite simple groups sectional rank and the ordinary rank (the biggest rank of an abelian subgroup) almost coincide as well? – Mark Sapir Jun 24 '12 at 5:08
Thank you very much for the answer. Is there something known about $ar(S)$ for finite simple groups? – Lior Bary-Soroker Jun 24 '12 at 6:28
@Mark: I am not sure. – Geoff Robinson Jun 24 '12 at 10:05
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. – Geoff Robinson Jun 24 '12 at 10:11

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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This does not answer the question. – Andreas Thom Jun 23 '12 at 12:38
@Andreas: Serves me right for not reading it properly. Thanks. – Brendan McKay Jun 24 '12 at 2:53
but interesting nonetheless! – oxeimon Feb 10 '15 at 23:04

Nobody seems to have mentioned the work of Burness, Liebeck and Shalev yet:

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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