= Solution
A <group extension> of $G$ by the $\mathbb ZG$-module $M$ is an exact sequence
$$
1\longrightarrow M\xrightarrow{i}E\xrightarrow{\pi}G\longrightarrow1
$$
whose 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-cocycle>
$$
c(g,h)=s(g)s(h)s(gh)^{-1}\in M.
$$
Associativity 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 $G=\langle x,y\rangle\cong\mathbb Z^2$ and let $I$ be the <augmentation ideal> of $\mathbb ZG$. The <Koszul resolution for a rank-two free abelian group> gives, after applying $\operatorname{Hom}_{\mathbb ZG}(-,M)$, the last coboundary
$$
M^2\longrightarrow M,
\qquad (a,b)\longmapsto(1-y)a+(x-1)b.
$$
Its image is $IM$. For $M=\mathbb ZG/I^2$ this yields the <second cohomology of a rank-two free abelian group with truncated group-ring coefficients> calculation
$$
H^2(G,M)=M/IM
\cong(\mathbb ZG/I^2)/(I/I^2)
\cong\mathbb ZG/I
\cong\mathbb Z.
$$
The 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 $F_2=\langle x,y\rangle$ 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 $\mathbb ZG$ on $[x,y]$ and is isomorphic to $\mathbb ZG/I^2$. Quotienting it by $I/I^2$ gives the central kernel $\langle[x,y]\rangle\cong\mathbb Z$ of the Heisenberg group. The kernel of
$$
N\longrightarrow F_2/\gamma_3(F_2)
$$
is $\gamma_3(F_2)/\gamma_4(F_2)$, 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.
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