= Solution
Write the <Heisenberg group> element as $M(x,y,z)$. Matrix multiplication and inversion give
$$
M(x,y,z)M(x',y',z')=M(x+x',\ y+y'+xz',\ z+z'),\qquad
M(x,y,z)^{-1}=M(-x,-y+xz,-z).
$$
All matrices have <determinant> one. They contain $M(0,0,0)=I$, are closed under multiplication and inverses, and inherit associativity from <matrix multiplication>; hence they form a <subgroup> of $GL_3(\mathbb R)$.
Two such matrices commute precisely when $xz'=x'z$. Requiring this for every $x',z'$ forces $x=z=0$. Consequently
$$
\boxed{Z(H)=\{M(0,y,0):y\in\mathbb R\}.}
$$
The map $\pi(M(x,y,z))=(x,z)$ is a surjective <group homomorphism> to the additive <group> $\mathbb R^2$, with <kernel of a group homomorphism> $Z(H)$. Equivalently, it gives the explicit <quotient group> isomorphism
$$
\boxed{H/Z(H)\longrightarrow(\mathbb R^2,+),\qquad M(x,y,z)Z(H)\longmapsto(x,z).}
$$
It is well-defined because multiplying by a central element changes only $y$. It is injective because equal $(x,z)$ imply that the quotient of the two matrices is central, and surjectivity follows by taking $y=0$. This also proves the needed instance of the <first isomorphism theorem> directly.
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