Abelian subgroup of PSL(3,q) Hello 
I know that there are abelian subgroups of order (q^2-1)/d (cyclic group) and (q-1)^2/d and q^2(elementary abelian) and (q-1)q/d in the group PSL(3,q), where d=(3,q-1).
Also I saw in a text that 
1) The orders of maximal abelian subgroups which divide (q-1)^2(q+1) are equal to (q-1)^2/d.
and 2) The orders of maximal abelian subgroups which divide q^2(q-1)^2 are equal  to q(q-1)/d.
and 3) The orders of maximal abelian subgroups which divide q^3 are equal to q^2.
But I don't know the reason of the above results,  could you kindly help me and let me know why the above facts are true?
Best regards Darya
 A: Here's a rough answer as I have very little time. I'm going to work in $K=SL(3,q)$ rather than $PSL(3,q)$ as it amounts to the same thing.
(1) Observe first that the centralizer of a $p$-element of $K$ has order $q^3$ or $q(q-1)$. This is just a matter of working with matrices. Then you can check that the centralizer of order $q(q-1)$ is abelian and you've got your first maximal. The only other possibility, then, is that a maximal abelian with order dividing $p$ has order dividing $q^3$. i.e. it lies in a Sylow $p$-subgroup, so is conjugate to a bunch of upper-triangular matrices. Now work with matrices to convince yourself that the maximal order is $q^2$.
(2) So we are left with abelian group of order dividing $(q-1)^2(q+1)$. You can use the theory of Jordan rational forms here: if a matrix has eigenvalues that don't lie in $\mathbb{F}_q$ then its centralizer has order $q^2-1$ and is abelian, hence maximal abelian. If all eigenvalues of all elements of your abelian group lie in $\mathbb{F}_q$, then they are simultaneously diagonalizable and so a maximal such group has order $(q-1)^2$.
