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The identity belongs to the center of a group. If $a,b$ are central, then for every $g\in G$,so the center is a subgroup. For a central $a$, $ga{g}^{-1}=a$ for every $g$, giving $gZ(G)g^{-1}=Z(G)$. Thus the center is a normal subgroup.

Write a matrix of the Heisenberg group as $h(x,y,z)$. Direct matrix multiplication and inversion giveThe identity is $h(0,0,0)$ and each matrix has determinant one, so these formulas show closure under products and inverses inside $G{L}_3(\mathbb{R})$. 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. ThusDefine the surjective group homomorphism $F:H\to ({\mathbb{R}}^2,+)$ by $F(h(x,y,z))=(x,z)$. The multiplication formula gives $F(h{h}^{\prime })=F(h)+F(h^{\prime })$, and its kernel of a group homomorphism is precisely $Z(H)$. The first isomorphism theorem yieldsThis 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.