Center of SU(3)

Question

I assumed a 3x3 matrix of the form $$A= \begin{pmatrix} a & b & c\\ d & e & f\\ k & l & m \end{pmatrix}$$ Then, since we know that the center is always an Abelian invariant subgroup and AB=BA, and the Gell-Mann matrices belong in SU(3), i took $$B=\begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$$ and $$B=\begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 0 \end{pmatrix}$$ which made the first matrix $$A=\begin{pmatrix} a & 0 & 0\\ 0 & a & 0\\ 0 & 0 & m \end{pmatrix}$$ Knowing that $$det(A) =1$$ led to $$a^2m=1$$ And that's my question. Shouldn't i get numbers and not a and m for the center of the group?

Answer 1

The center of $SU(3)$ is isomorphic to the cyclic group $C_3$, generated by $$ e^{2\pi i/3}I_3. $$ You have written down elements of the Lie algebra $\mathfrak{su}(3)$, which are elements of a vector space.