I'm reading an article https://webusers.imj-prg.fr/~julie.deserti/biblio/Blanc_linearisationoffiniteabeliansubgroupsoftheCremonagroupoftheplane.pdf and I have problems with understanding the proof of Proposition 2.2. One of the statements of the proof is that (here we denote a morphism $[x:y:z]\to [ax:by:cz]$ by $[a:b:c]$):
"A simple calculation shows that every finite abelian subgroup of $Aut(\mathbb{P}_{\mathbb{C}}^2)=PGL(3,\mathbb{C})$ is either diagonalizable or conjugate to the group $V_9$ (here $V_9\cong \mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}$ and is generated by $[1:\xi_3:\xi_3^2]$ and $[x:y:z]\to [y:z:x]$ where $\xi_3=e^{2i\pi/3}$)"
I tried my best to prove it with the direct calculation, but I'm afraid that I'm doing something wrong. I will appreciate any help!
Edit: (details of my attempt):
First, I took a matrix $g\in PGL(3,\mathbb{C})$ and assumed that it is diagnosable. We can assume $a_{11}$ so it will be of the form $$g=\begin{pmatrix} 1 &0 & 0\\ 0 & \alpha &0 \\ 0 & 0 & \beta \end{pmatrix} $$
Then since the group is finite (so for any $g\in G$ we have that $g^{m+1}=g$ fo some $g$) we get that $\alpha=\xi_m^a$, $\beta=\xi_m^b$ (where $\xi_{m}=e^{2\pi i/m}$). Then if $a\ne b$ the we can take any other element $h\in G$ of the form:
$$h=\begin{pmatrix} 1 & h_{12} & h_{13}\\ h_{21} & h_{22} & h_{23} \\ h_{31} & h_{32} & h_{33} \end{pmatrix} $$
and from the fact that $gh=hg$ (looking at the explicit expansions) we obtain that $h_{12}=h_{13}=h_{23}=h_{21}=h_{31}=h_{32}=0$ so the matrix is diagonisable:
$$\begin{pmatrix} 1 & 0 & 0\\ 0 & h_{22} & 0 \\ 0 & 0 & h_{33} \end{pmatrix} $$
and similarly to the above since the group is finite we get that $h_{22}=\xi_n^c$, $h_{33}=\xi_n^d$ (where $\xi_{n}=e^{2\pi i/n}$).
In case $a=b$ (or $a=0$ or $b=0$ it seems that they lead to the similar cases) other elements $h$ are of the form:
$$\begin{pmatrix} 1 & 0 & 0\\ 0 & A & B \\ 0 & C & D \end{pmatrix} $$
and I couldn't get more information for this case.
Moreover, I do not see at all how to get $V_9$ (and how to get "not diagonalizable element"$[x:y:z]\to [y:z:x]$) from the direct calculations + I cannot just assume that any element I take will be diagonalizable. I also tried multiplying 3x3 matrixes directly but got lost in these calculations.
I really tried, but I failed and it seems that my way does not work. I will appreciate any help!