In previous questions, I've proved that $B$ is a free abelian group of rank $2$. Then naturally $B$ is isomorphic to $\Bbb Z^2$, right? Then $\mathbb{Z}^2/B$ is isomorphic to $\mathbb{Z}^2/\mathbb{Z}^2$.
Isn't this just the trivial group?
How am I suppose to express it as a direct product?
Observe that $\Bbb Z\cong 2\Bbb Z\cong 3\Bbb Z$.
Let $G=\langle a,b\mid ab=ba\rangle\cong \Bbb Z^2$.
Suppose
$$\begin{align} B&:=\langle a^2,b^2\rangle_G \\ &\cong 2\Bbb Z\times 2\Bbb Z\\ &\cong \Bbb Z^2. \end{align}$$
Then
$$\begin{align} G/B&\cong \langle a,b\mid ab=ba\rangle/\langle a^2,b^2\rangle_G\\ &=\langle a,b\mid a^2, b^2, ab=ba\rangle\\ &\cong \Bbb Z_2\times \Bbb Z_2. \end{align}$$
But if
$$\begin{align} H&:=\langle a^2,b^3\rangle_G \\ &\cong 2\Bbb Z\times 3\Bbb Z\\ &\cong \Bbb Z^2, \end{align}$$
then
$$\begin{align} G/H&\cong \langle a,b\mid ab=ba\rangle/\langle a^2,b^3\rangle_G\\ &=\langle a,b\mid a^2, b^3, ab=ba\rangle\\ &\cong \Bbb Z_2\times \Bbb Z_3. \end{align}$$