In General linear group $GL(n,q)$, every matrix is conjugate with its rational canonical form. Using this fact we have the conjugacy classes of cyclic groups. My question is about the groups which are generated by two elements. I tried to find some results about conjugacy classes but it seems to be hard, any comment is so appreciated.

2$\begingroup$ I am doubtful whether you will find much in the literature about conjugacy classes of $2$generator subgroups of ${\rm GL}(n,q)$ in general. There are just too many of them, and the $2$generator condition does not provide enough restriction to get a handle on them. Indeed, deciding whether a particular subgroup is $2$generated is not always easy. There is a lot of information known about maximal subgroups of ${\rm GL}(n,q)$, particularly for small values of $n$. $\endgroup$ – Derek Holt Feb 20 '14 at 8:37

$\begingroup$ I tried to restrict the problem to consider all $2$generator subgroups $G$ of $GL(n,q)$ which $G=<A><B>$, where by $<A>$ we mean a cyclic group. But it seems it is complicated too. do you have any comment? $\endgroup$ – user33209 Feb 20 '14 at 9:07

1$\begingroup$ One thing we have to contend with is the fact that every finite simple group can be generated by two elements. ( arxiv.org/abs/1009.6183 ) $\endgroup$ – P Vanchinathan Feb 20 '14 at 9:51

$\begingroup$ Please change "to" into "two" in your title. $\endgroup$ – P Vanchinathan Feb 20 '14 at 9:52
You would have an answer to your question if you could classify pairs of elements in $GL_n(\mathbf F_q)$ up to simultaneous conjugacy. This is the notorious matrix pair problem, which is the quintessential wild problem in representation theory (see When is a classification problem "wild"? , Are wild problems related to undecidable ones? and How can classifying irreducible representations be a "wild" problem?). So you are unlikely to get a nice answer for general $n$.

1$\begingroup$ In particular the first order theory of an F_q vector space with two automorphisms is undecidable. $\endgroup$ – Benjamin Steinberg Feb 20 '14 at 14:00