Thanks for any help or comments.
Is it possible to recognize all maximal abelian subgroups of general linear group on finite field $F$ of order $q$, $GL_n(F)$. By maximal abelian I mean if $A$ is maximal abelian and $B$ is abelian such that $A\subseteq B$, then $B=A$.
This must be well documented, but I think the systematic way to do this is to proceed by induction on $n$. Note that if $A$ is a maximal Abelian subgroup of $G = {\rm GL}(n,q)$, then $A$ is a maximal Abelian subgroup of $C_{G}(a)$ for each $a \in A^{\#}.$ If $A$ contain a semisimple element $a$ which does not act indecomposably on the natural module, then we can reduce the question to one about maximal Abelian subgroups of smaller general linear groups ( possibly over larger fields) as follows: If the characteristic polynomial of $a$ has two different irreducible factors, then $A$ is contained in a direct product of smaller dimensional linear groups, and hence (by maximality) is the direct product of maximal Abelian subgroups of those general linear groups. If the characteristic polynomial of $a$ is a power of a single irreducible of degree greater than one, then $C_{G}(a)$ is isomorphic to a group of the form ${\rm GL}(d,q^{s})$ for some $d < n.$ Hence we reduce to the case that every semisimple element of $A$ is a scalar matrix. This reduces us (up to conjugacy) to the problem finding the maximal Abelian subgroups of the group of unipotent upper triangular matrices. I am not sure how difficult this is, though at first sight it does not seem easy to me.
Questions of this type have been raised about various finite groups of Lie type at MathOverflow previously, for example here. As Nick Gill's comment indicates, the work of E. Vvodin is worth consulting, along with an earlier paper by M. Barry, etc. Naturally the general (or special) linear group over a finite field is somewhat easier to study directly, using a mixture of techniques from linear algebra and finite group theory. But there is some advantage in looking at all finite groups of Lie type from the perspective of algebraic groups.
Two basic questions tend to arise: (1) determine (up to conjugacy) all maximal abelian subgroups, (2) find the largest order of any such subgroup. From either the linear algebra or the algebraic group viewpoint, a natural tool here is the Jordan decomposition of elements. It turns out that semisimple elements (those of order not divisible by $p$) are the easiest to study systematically, largely because their centralizers are again reductive -- and even connected when the algebraic group is simply connected (true here for $\mathrm{SL}_n$).
In particular, an abelian subgroup $A$ of the finite group $G$ consisting of semisimple elements always lies in a maximal torus of the algebraic group defined over $\mathbb{F}_q$. This is one of the results developed for all finite Chevalley groups by Springer and Steinberg in their extensive notes on conjugacy classes: Part E in Seminar on Algebraic Groups and Related Finite Groups (Springer Lecture Notes in Math. 131, 1970), II, 5.8-5.12. (But there are nuances for some primes in types other than the special linear groups.) In particular, the groups of rational points (or fixed points under Steinberg's endomorphism $\sigma$) in the various maximal tori are easily seen to be maximal abelian subgroups and have orders specified in terms of data from the Weyl group, here $S_n$: II, 1.7. These orders are approximately $q^n$ for the finite general linear groups ($n$ being the overall rank). Inductive methods like those suggested by Geoff Robinson are often helpful when only semisimple elements are discussed.
The complication is that the maximum order of an abelian subgroup is approximately $q^{n^2/4}$, typically much larger than a finite torus. Since the centralizers in the algebraic group of nontrivial unipotent elements (= elements having $p$-power orders) are usually far from being reductive, it is tricky to work out the orders of all maximal abelian subgroups of $G$ which involve such elements.
