most recent 30 from http://mathoverflow.net 2013-05-25T06:15:05Z http://mathoverflow.net/feeds/question/100446 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100446/number-of-generators-of-a-subgroup-of-a-finite-simple-group Lior Bary-Soroker 2012-06-23T11:45:28Z 2012-11-26T07:17:53Z

<p>For a finite group $G$ we denote $d(G)$ the minimal size of a set of generators of $G$. We define $D(G) = \max( d(H) \mid H\leq G)$. </p> <p>Let $S$ be a finite simple group. Are there `good' bounds on $D(S)$ in terms of the size of $S$? </p>

http://mathoverflow.net/questions/100446/number-of-generators-of-a-subgroup-of-a-finite-simple-group/100447#100447 Brendan McKay 2012-06-23T11:51:12Z 2012-06-23T11:51:12Z

<p>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 <a href="http://www.springerlink.com/content/r32065tp23lu7847/" rel="nofollow">here</a> for example.</p>

http://mathoverflow.net/questions/100446/number-of-generators-of-a-subgroup-of-a-finite-simple-group/100489#100489 Geoff Robinson 2012-06-23T22:22:28Z 2012-07-30T23:16:27Z

<p>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.$</p>

http://mathoverflow.net/questions/100446/number-of-generators-of-a-subgroup-of-a-finite-simple-group/106512#106512 Ben Fairbairn 2012-09-06T14:11:00Z 2012-09-06T14:11:00Z

<p>Nobody seems to have mentioned the work of Burness, Liebeck and Shalev yet:</p> <p><a href="http://www.personal.soton.ac.uk/tb1u06/docs/maxgen26.pdf" rel="nofollow">http://www.personal.soton.ac.uk/tb1u06/docs/maxgen26.pdf</a></p> <p>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.</p>