TITLE: Is every finite $d$-dimensional matrix group generated by $d$ elements?
QUESTION [13 upvotes]: The question is in the title. If $\Gamma\subset\mathrm{GL}(\Bbb R^d)$ is a finite matrix group, can it be generated by (at most) $d$ elements?
I suspect that this hope is too naive, but I have no counterexamples. I would also be interested in bounds on the number of generators.

REPLY [19 votes]: The answer to the question is yes.
In Theorem 1.2 of this paper. the authors (Colva Roney-Dougal and myself) prove that  if $G \le {\rm GL}(n,F)$ with $F$ a field, where
(a) $G$ is finite;
(b) either $G$ is completely reducible or ${\rm char} F = p$ and $O_p(G)=1$; and
(c) $F$ does not contain a primitive fourth root of unity,
then $d(G) \le n$.
This is best possible when ${\rm char} F \ne 2$,  because there is an obvious embedding of $C_2^n$ in ${\rm GL}(n,F)$.