The discrete Heisenberg group has growth degree four

Matrix multiplication gives $(a,b,c)(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})=(a+a^{\mathrm{\prime }},b+b^{\mathrm{\prime }},c+c^{\mathrm{\prime }}+a{b}^{\mathrm{\prime }})$. The identity is (0,0,0), and substitution on both sides gives $(a,b,c{)}^{-1}=(-a,-b,-c+ab)$. Put $x=(1,0,0)$, $y=(0,1,0)$, $z=(0,0,1)$. Their integer powers satisfy $x^a{y}^b{z}^k=(a,b,ab+k)$, so every matrix is uniquely $x^a{y}^b{z}^{c-ab}$.

Using the product and inverse formulas gives $[(a,b,c),(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})]=(0,0,a{b}^{\mathrm{\prime }}-a^{\mathrm{\prime }}b)$. In particular $[x,y]=z$. Every commutator is a power of z and z itself is a commutator, so $[H,H]=⟨z⟩$. Elements (0,0,c) commute with every triple by the product rule, hence ${\gamma }_3=1$. Conversely, if $(a,b,c)$ commutes with $x$ and $y$, its commutators have central entries $-b$ and $a$, so $a=b=0$. Thus the center is exactly $⟨z⟩$. Since z has infinite order, H has class exactly two. The elements x,y generate because z=[x,y] and step 1.1 gives every matrix.

The homomorphism $(a,b,c)\mapsto (a,b)$ is onto ${\mathbb{Z}}^2$ with kernel $⟨z⟩$, while $c\mapsto (0,0,c)$ identifies the kernel with $\mathbb{Z}$. Thus $r_1=2$ and $r_2=1$, giving $D=1\cdot 2+2\cdot 1=4$. F1 applies because x,y are finite generators and H has class two. The identity has a=b=k=0; the matrix entry c equals ab+k and is not generally the normal coordinate k.