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Take $G=S_3$ and its normal subgroup $N=A_3=⟨r⟩$, where $r=(123)$. Normality follows either from the sign of a permutation or directly from conjugation sending $r$ to $r$ or $r^{-1}$. The quotient group has order two, and $t=(12)$ provides a section $i:C_2\to S_3$ because $t^2=1$ and $tN$ is its nonidentity coset.

However, the map $\Phi (q,n)=i(q)n$ from the direct product of groups is not a group homomorphism. In its domain $(1,r)$ commutes with $(tN,1)$, whereas their images $r$ and $t$ do not commute: $tr{t}^{-1}=r^{-1}\ne r$. This is a split group extension giving a genuine semidirect product instead of a direct product. The pair $S_3,A_3$ supplies the required split but non-direct example.

For the general bijectivity assertion, let $\pi :G\to G/N$ be the quotient projection and assume $\pi \circ i$ is the identity. Given $g\in G$, put $q=\pi (g)$ and $n=i(q{)}^{-1}g$. Then $\pi (n)=q^{-1}q=1$, so $n\in N$ and $g=i(q)n$. This proves surjectivity of $\Phi$. If $i(q)n=i(q^{\prime })n^{\prime }$, applying $\pi$ gives $q=q^{\prime }$, after which cancellation gives $n=n^{\prime }$. ThusThis section normal form for a split group extension does not require finiteness. The obstruction to its being a homomorphism is the nontrivial conjugation action of the section on $N$.

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.