Number of generators of a subgroup of a finite simple group 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 $S$? 
 A: 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.
A: 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.$
A: 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.
