I am trying to solve the following problem:
Let $G$ be a finite group such that there exists an isomorphism $\phi:f:G/Z(G) \to \mathbb Z_3 \oplus \mathbb Z_3$ and $x \in G$ such that $f(\overline{x})=(1,0)$. Show that for $y \in G$, $$xy=yx \leftrightarrow f(\overline{y}) \in \langle (1,0) \rangle$$
I could prove the left implication, if $f(\overline{y}) \in \langle (1,0) \rangle$, then $f(\overline{y})=kf(\overline{x})$ for $k=0,1,2$. From here we have the cases $y,yx^{-1},yx^{-2} \in Z(G)$ and from here I could prove that $yxy^{-1}x^{-1}=e$.
I don't know what to do to show the other implication. I know $f(\overline{y})=k(1,0)+t(0,1)$ since $\mathbb Z_3 \oplus \mathbb Z_3=\langle (1,0),(0,1) \rangle$ but I don't know how to deduce that $t=0$. Any hints would be greatly appreciated.