Central extensions of a cyclic group

In any central extension, choose a lift $\tilde{x}$ of the generator $x$. Since the quotient has order $n$, the element ${\tilde{x}}^n$ lies in the central kernel $A$; write it as $a\in A$. For every $a\in A$, construct $E_a=(A\times \mathbb{Z})\mathrm{/}⟨(-a,n)⟩\mathrm{.}$ The subgroup in the denominator is normal because $A\times \mathbb{Z}$ is abelian. The map $A\to E_a$, $b\mapsto [(b,0)]$, is injective: if $(b,0)=k(-a,n)$ then $kn=0$, hence $k=0$ and $b=0$. The map $E_a\to C_n$, $[(b,m)]\mapsto x^m$, is well defined and surjective, with kernel precisely the image of $A$; if $m=kn$, then $[(b,m)]=[(b+ka,0)]$. Thus $E_a$ is a central extension, and its distinguished lift $[(0,1)]$ has $n$th power $[(a,0)]$.

The homomorphism $A\times \mathbb{Z}\to E$ defined by $(b,m)\mapsto b{\tilde{x}}^m$ is onto: an element of $E$ has quotient $x^m$, so differs from ${\tilde{x}}^m$ by an element of $A$. Its kernel is exactly $⟨(-a,n)⟩$: a kernel element has $m=kn$, and then $b{\tilde{x}}^{kn}=b+ka=0$ in the embedded kernel. Therefore it descends to an extension equivalence $E_a\cong E$. Every central extension in the example is realized by one of the $E_a$.

Replacing $\tilde{x}$ by $b\tilde{x}$ changes its $n$th power from $a$ to $a+nb$ because $A$ is central. Conversely, if $a^{\mathrm{\prime }}=a+nb$, sending the distinguished lift $[(0,1)]\in E_{a^{\mathrm{\prime }}}$ to $[(b,1)]\in E_a$ and fixing $A$ defines an extension equivalence: the new lift has $n$th power $a+nb=a^{\mathrm{\prime }}$. Hence two parameters define equivalent extensions exactly when they differ by an element of $nA$.

Steps 1.1--3.1 identify the extension classes with $A\mathrm{/}nA$. Under the stated Choice premise, [L1] identifies those same classes with $H^2(C_n,A)$, giving the claimed cohomological description.