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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 $S$?

edited Nov 26 at 7:17

asked Jun 23 2012 at 11:45

Lior Bary-Soroker
1,355●7●18

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 2012 at 13:08

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$\log_2 |S|$ 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 2012 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 2012 at 13:56

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No it seems that $A_n$ is better asymptotically ($n!$ vs $q^{n^2}$). – Mark Sapir Jun 23 2012 at 14:24

by mistake I rolled back, and can't find how to undo. HELP!!! – Lior Bary-Soroker Nov 25 at 10:12

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Geoff Robinson · Accepted Answer · 2012-07-30 23:16:27Z UTC

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 $P$) f 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 number of generators of any section of $G$ which is a $p$-group (a section of $G$ is a group of the form $H/K$ where $H$ is a subgroup of $G$ an $K \lhd H ).$ 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.$

Ben Fairbairn · Answer 2 · 2012-09-06 14:11:00Z UTC

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

http://www.personal.soton.ac.uk/tb1u06/docs/maxgen26.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.

Brendan McKay · Answer 3 · 2012-06-23 11:51:12Z UTC

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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.

answered Jun 23 2012 at 11:51

Brendan McKay
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This does not answer the question. – Andreas Thom Jun 23 2012 at 12:38

@Andreas: Serves me right for not reading it properly. Thanks. – Brendan McKay Jun 24 at 2:53

Number of generators of a subgroup of a finite simple group

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