= Solution
The identity belongs to the <center of a group>. If $a,b$ are central, then for every $g\in G$,
$$
g(ab^{-1})=a\,gb^{-1}=ab^{-1}g,
$$
so the center is a <subgroup>. For a central $a$, $gag^{-1}=a$ for every $g$, giving $gZ(G)g^{-1}=Z(G)$. Thus \b[the center is a <normal subgroup>].
Write a <matrix> of the <Heisenberg group> as $h(x,y,z)$. Direct <matrix multiplication> and inversion give
$$
h(x,y,z)h(a,b,c)=h(x+a,y+b+xc,z+c),
\qquad
h(x,y,z)^{-1}=h(-x,xz-y,-z).
$$
The identity is $h(0,0,0)$ and each <matrix> has <determinant> one, so these formulas show closure under products and inverses inside $GL_3(\mathbb R)$. Hence $H$ is a <subgroup>.
The two products $h(x,y,z)h(a,b,c)$ and $h(a,b,c)h(x,y,z)$ differ only in their upper-right entries, by $xc-az$. Commutation with every $h(a,b,c)$ therefore requires $xc=az$ for all real $a,c$. Setting $a=0,c=1$ forces $x=0$; setting $a=1,c=0$ forces $z=0$. These conditions are also sufficient. Thus
$$
\boxed{Z(H)=\{h(0,y,0):y\in\mathbb R\}.}
$$
Define the surjective <group homomorphism> $F:H\to(\mathbb R^2,+)$ by $F(h(x,y,z))=(x,z)$. The multiplication formula gives $F(hh')=F(h)+F(h')$, and its <kernel of a group homomorphism> is precisely $Z(H)$. The <first isomorphism theorem> yields
$$
\boxed{H/Z(H)\cong(\mathbb R^2,+),\qquad
h(x,y,z)Z(H)\longmapsto(x,z).}
$$
This <center quotient of the real Heisenberg group> can also be checked directly: two elements have the same $(x,z)$ coordinates exactly when they differ by multiplication by a central element, and quotient multiplication adds these coordinates.
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