= Solution
Write the matrix as $M(a,b,x)$. Direct multiplication and inversion give
$$
M(a,b,x)M(a',b',x')=M(a+a',b+b',x+x'+ab'),\qquad
M(a,b,x)^{-1}=M(-a,-b,ab-x).
$$
The identity is $M(0,0,0)$ and all determinants are one, so these formulas verify the <subgroup> criterion inside $\mathrm{GL}_3(\mathbb R)$. This is the <real Heisenberg group>.
The map $M(a,b,x)\mapsto a$ is a surjective <group homomorphism> to the additive <group> of real numbers; its kernel is exactly $H$. Hence $H$ is normal and the <first isomorphism theorem> gives
$$
\boxed{G/H\cong(\mathbb R,+).}
$$
Two matrices commute exactly when $ab'=a'b$. Requiring that relation for every $a',b'$ forces $a=b=0$, while $x$ is unrestricted. Therefore
$$
\boxed{Z(G)=\{M(0,0,x):x\in\mathbb R\}.}
$$
Finally $M(a,b,x)\mapsto(a,b)$ is a surjective <group homomorphism> to the additive <group> $\mathbb R^2$ with precisely this <kernel of a group homomorphism>, giving \b[$G/Z(G)\cong(\mathbb R^2,+)$]. The central coordinate records noncommutativity: the <group commutator> is $M(0,0,ab'-a'b)$.
Back to article page
