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 $S$?
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 $S$? 


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 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 $G$ 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.$ 


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 nonabelian 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. 


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. 

