= Solution
Write an element of the <Integer Heisenberg group> as $(x,y,z)$. <Matrix multiplication> gives
$$
(x,y,z)(x',y',z')=(x+x',y+y',z+z'+xy').
$$
The subgroup
$$
N=\{(0,y,z):y,z\in\mathbb Z\}
$$
is normal and isomorphic to $\mathbb Z^2$. If $t=(1,0,0)$, then
$$
t(0,y,z)t^{-1}=(0,y,z+y).
$$
Thus, on the coordinate column $(y,z)^T$, <conjugation> by $t$ is the <linear map> with matrix
$$
A=\begin{pmatrix}1&0\\1&1\end{pmatrix}.
$$
Every element has a unique expression $(0,y,z)t^x$, so
$$
H\cong\mathbb Z^2\rtimes_A\mathbb Z.
$$
The commutators $[t,(0,y,z)]$ fill the central subgroup of matrices $(0,0,z)$, while the quotient by this subgroup is generated freely and abelianly by the images of $(1,0,0)$ and $(0,1,0)$. Equivalently, $(A-I)\mathbb Z^2$ is the second coordinate axis. Therefore the <abelianization> is
$$
H^{\mathrm{ab}}\cong
\mathbb Z\oplus\mathbb Z^2/(A-I)\mathbb Z^2
\cong\mathbb Z^2.
$$
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