Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $S$?

share|improve this question
    
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
3  
$\log_2 |S|$ 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
1  
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
show 2 more comments

3 Answers

up vote 9 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 $P$) 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.$

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
7  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.