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The quotient mapis a central nonsplit extension of the Integer Heisenberg group by ${\gamma }_3(F_2)/{\gamma }_4(F_2)\cong {\mathbb{Z}}^2$. It cannot split because the source has abelianization ${\mathbb{Z}}^2$, whereas a central splitting would give an abelianization containing an additional direct factor ${\mathbb{Z}}^2$.

Write the Integer Heisenberg group with central generator $c=[a,b]$, so $ab=cba$. Moving $a$ successively past powers of $b$ givesFor positive exponents this follows by two inductions; inverses give the remaining signs. The representation $a=I+E_{12}$, $b=I+E_{23}$, $c=I+E_{13}$ checks the identity and shows $c$ has infinite order.

The quotient of the real Heisenberg group by the Integer Heisenberg group is the mapping torus of the torus automorphismIts universal cover is the contractible real Heisenberg group, so its fundamental group is the integer Heisenberg group.

For the Integer Heisenberg group, choose $x=I+E_{12}$, $y=I+E_{23}$ and central $z=I+E_{13}$. Every element has a unique form $y^c{z}^b{x}^a$, and $xy=zyx$. The group algebra is therefore the displayed skew Laurent polynomial ring, with $\sigma (z)=z$. The base is a commutative Laurent polynomial ring and an integral domain; the skew Hilbert basis theorem gives both-sided Noetherianity and the leading coefficient of a product proves that there are no zero divisors.

The two commutation relations make $c$ central, and the remaining relation is $ab=cba$. Repeatedly move $a$ past $b$, keeping the central powers in front. Induction first on $s$ and then on $r$ givesThereforeThis is the commutator identity in the integer Heisenberg group. The order of the factors fixes the positive sign of the exponent.

To check the claimed infinite order, map the generators to matricesTheir commutator is $I+E_{13}$, which commutes with both other matrices, so all defining relations hold. The image of $c^m$ is $I+m{E}_{13}$; hence $c$ has infinite order. This realizes the presentation in the Integer Heisenberg group and is sufficient to certify the cyclic subgroup, without needing to prove faithfulness of the whole representation.

With the ambient word metric from $\{a,b,c{\}}^{\pm 1}$, the displayed commutator givesFor any finite generating set $T$ of $⟨c⟩$, write its elements as powers of $c$ and set $B=\max \{|j|:c^j\in T\}$, with inverses included. Since $T$ generates an infinite cyclic group, $B\ge 1$. A word of length $m$ in $T$ has exponent at most $Bm$ in absolute value, soA quasi-isometric embedding of the inclusion would require $|c^{k^2}{|}_H\ge {\lambda }^{-1}|c^{k^2}{|}_T-\epsilon$. Combining these bounds would give $4k\ge k^2/(B\lambda )-\epsilon$ for every $k$, which is impossible. Thus the inclusion of the central infinite cyclic subgroup is not a quasi-isometric embedding. Changing the ambient finite generating set only changes multiplicative constants, by equivalence of finite word metrics. This is the quadratic distortion of the Heisenberg center.

Write $v(x,y,z)$ for the displayed upper-unitriangular matrix. Matrix multiplication givesThus $V$ is the real Heisenberg group and $\Gamma$ is its Integer Heisenberg group. The subgroup $\Gamma$ is discrete. Choose an identity neighbourhood $U\subset V$ so small thatThen the translates $\gamma U$ are pairwise disjoint, and the quotient map restricts on each one to a homeomorphism onto the same open subset of $N$. Hence every point has an evenly covered neighbourhood andThis is the general quotient of a topological group by a discrete subgroup construction.

The Integer Heisenberg groupis generated by the matrices with $(a,b,c)=(1,0,0)$ and $(0,1,0)$. Their group commutator is the nonidentity central matrix with $(a,b,c)=(0,0,1)$. Thus it is nonabelian and nilpotent of class two, while being finitely generated.

Write an element of the Integer Heisenberg group as $(x,y,z)$. Matrix multiplication givesThe subgroupis normal and isomorphic to ${\mathbb{Z}}^2$. If $t=(1,0,0)$, thenThus, on the coordinate column $(y,z{)}^T$, conjugation by $t$ is the linear map with matrixEvery element has a unique expression $(0,y,z)t^x$, so

The commutators $[t,(0,y,z)]$ fill the central subgroup of matrices $(0,0,z)$, while the quotient by this subgroup is generated freely and abelianly by the images of $(1,0,0)$ and $(0,1,0)$. Equivalently, $(A-I){\mathbb{Z}}^2$ is the second coordinate axis. Therefore the abelianization is

A group extension of $G$ by the $\mathbb{Z}G$-module $M$ is an exact sequencewhose conjugation action on $i(M)$ agrees with the prescribed action of $G$ on $M$. It is a split group extension when $\pi$ has a group-homomorphic section $s:G\to E$. Two such extensions are equivalent group extensions when an isomorphism of their middle groups is the identity on $M$ and induces the identity on $G$. Transporting a section through that isomorphism proves that every extension equivalent to a split extension is split.

Choose a set-theoretic section $s$ with $s(1)=1$. Its failure to preserve multiplication is the normalized two-cocycleAssociativity gives the two-cocycle identity, and replacing $s$ changes $c$ by a group coboundary. The resulting class $[c]\in H^2(G,M)$ is therefore intrinsic to the extension, as expressed by second group cohomology classifies group extensions.

Now write $G=⟨x,y⟩\cong {\mathbb{Z}}^2$ and let $I$ be the augmentation ideal of $\mathbb{Z}G$. The Koszul resolution for a rank-two free abelian group gives, after applying ${\mathrm{Hom}}_{\mathbb{Z}G}(-,M)$, the last coboundaryIts image is $IM$. For $M=\mathbb{Z}G/I^2$ this yields the second cohomology of a rank-two free abelian group with truncated group-ring coefficients calculationThe canonical map $M\to \mathbb{Z}$ induces the identity on these final quotients, so $H^2(G,M)\to H^2(G,\mathbb{Z})$ is surjective; indeed it is an isomorphism.

Let $F_2=⟨x,y⟩$ and let ${\gamma }_i(F_2)$ be its lower central series. The class-two quotient $F_2/{\gamma }_3(F_2)$ is the Integer Heisenberg group. In the class-three free nilpotent group $N=F_2/{\gamma }_4(F_2)$, the module ${\gamma }_2(F_2)/{\gamma }_4(F_2)$ is cyclic over $\mathbb{Z}G$ on $[x,y]$ and is isomorphic to $\mathbb{Z}G/I^2$. Quotienting it by $I/I^2$ gives the central kernel $⟨[x,y]⟩\cong \mathbb{Z}$ of the Heisenberg group. The kernel ofis ${\gamma }_3(F_2)/{\gamma }_4(F_2)$, freely generated by $[[x,y],x]$ and $[[x,y],y]$, and is central. Thus it is ${\mathbb{Z}}^2$. This is the central nonsplit extension of the integer Heisenberg group by ${\mathbb{Z}}^2$. If it split, centrality would give $N\cong H(\mathbb{Z})\times {\mathbb{Z}}^2$, whose abelianization has rank four; but $N$ has abelianization ${\mathbb{Z}}^2$. Hence the extension is nonsplit.

The central infinite cyclic subgroup of the Integer Heisenberg group has quadratic subgroup distortion. The identity $c^{k^2}=[a^k,b^k]$ gives ambient length at most $4k$ against intrinsic length $k^2$. For general $m\ge 0$, take $k=\lfloor \sqrt{m}\rfloor$ and write $m=k^2+r$, $0\le r\le 2k$, to obtain length at most $4k+r\le 6\sqrt{m}$. Conversely, a word of length $n$ in $a,b,c$ changes the first coordinate by at most $n$ and its central coordinate by at most $n^2+n$, so $|c^m|$ is bounded below by a positive multiple of $\sqrt{|m|}$ for $m\ne 0$. Negative exponents follow by inverses. Thus the ambient and intrinsic metrics have quadratic relative growth.