Maximal Abelian Subgroup of SL(3,C)

Question

So I'm looking to find the maximal abelian subgroup of SL(3,C). I know that if a maximal torus for SL(3,C) exists that said torus is the maximal abelian subgroup. Is it enough to know that since SU(3) is a subgroup that they have the same maximally abelian subgroup (namely, a maximal torus of SU(3))? Is there a simple way to go about showing that they share this maximal abelian subgroup?

Apologies if the wording lacks precision, feel free to guide me toward clarifications.

Answer 1

Surely the subgroup of diagonal matrices with determinant 1, and the subgroup of matrices of the form $\left(\begin{array}{ccc}a&b&c\\0&a&0\\0&0&a\end{array}\right)$ with $a^3=1$ are both self-centralizing, and hence they are maximal abelian subgroups of ${\rm SL}_3(\mathbb{C})$?

Added later: two more maximal abelian subgroups are matrices of the form $\left(\begin{array}{ccc}a&b&c\\0&a&b\\0&0&a\end{array}\right)$ with $a^3=1$, and matrices of form $\left(\begin{array}{ccc}a&b&0\\0&a&0\\0&0&a^{-2}\end{array}\right)$ with $a \ne 0$.

That makes at least four distinct conjugacy classes of maximal abelian subgroups. Perhaps they are the only ones.

Answer 2

The Lie group $SL_{3}\mathbb{C}$ is simply connected, thus the maximal torus is in one to one correspondence with the Cartan subalgebra of $sl_{3}\mathbb{C}$. We can identify it as generated by $e_{1,1}-e_{2,2}$, $e_{2,2}-e_{3,3}$. This coincide with the Cartan subalgebra of $su_{3}$. Thus I believe the maximal torus of them are the same.

Another (possibly wrong) way of thinking is the maximal torus of $SL_{3}\mathbb{C}$ has rank $2$ because it has two generators(the third uniquely determined by the previous two since it is traceless). We know for $SU_{3}$ the maximal torus is the intersection of $U_{n}$ with $T^{n}$, which cuts dimension by 1. Thus both maximal torus has rank 2. Since $SU_{3}\subset SL_{3}\mathbb{C}$ the two maximal torus should be conjugate to each other or can be identified to be the same.