Second Cohomology and Abelian Kernel Extensions - Examples

In $\mathbb{Z}\mathrm{/}{p}^2\mathbb{Z}$, adding two chosen lifts $i$ and $j$ either stays below $p$ or crosses the first multiple of $p$. Therefore $s(i)+s(j)-s(i+j<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p)=pf(i,j),$ with $f(i,j)$ given by the carry function above.

The kernel is central, so [L1] identifies the carry function with the extension class. If $[f]=0$, [L2] gives a normalized cochain $b:C_p\to C_p$ with $f=\delta b$. Subtract the kernel element $pb(i)$ from each chosen lift: $s^{\mathrm{\prime }}(i)=s(i)-pb(i)\in C_{p^2}$. Its factor set is $f-\delta b=0$, so $s^{\mathrm{\prime }}$ is a group-homomorphic section of the quotient. But $s^{\mathrm{\prime }}(1)$ would have order dividing $p$ and map to $1\in C_p$; every element of $C_{p^2}$ killed by $p$ lies in the kernel $p{C}_{p^2}$ and maps to $0$. This contradiction shows $[f]\mathrm{\ne }0$ without any global Choice premise.

Hence the displayed cocycle represents a nonzero class in $H^2(C_p,C_p)$.

With $f=0$, [F1] becomes $(m,g)(n,h)=(m+g\cdot n,gh),$ which is the usual semidirect-product law on $M⋊G$.

The explicit normalized section $s(g)=(0,g)$ is a homomorphism, so the extension splits. Its factor set is identically zero because $s(g)s(h)=s(gh)$; [L1] assigns the extension the class $[0]$, which is the zero class by [F2]. This uses the displayed section and needs no choice of fibres.

Therefore the zero cocycle gives the split extension.

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.

In both $D_8$ and $Q_8$, the center has order $2$, and quotienting by it gives $C_2\times C_2$. So each group defines a central extension of $C_2\times C_2$ by $C_2$.

The two extensions are not equivalent because $D_8$ contains five involutions, while $Q_8$ contains only one. An extension equivalence would be an isomorphism of middle groups preserving the kernel and quotient data, but the middle groups are not isomorphic.

Therefore [L1] assigns distinct classes in $H^2(C_2\times C_2,C_2)$ to the dihedral and quaternion central extensions.

For the standard section $s(t)=(0,t)$, the equality in [F1] gives $f_s=0$ because $(0,t{)}^2=(0,1)$.

Replace the section by $s^{\mathrm{\prime }}(t)=(1,t)$. Then $(1,t{)}^2=(2,1)$, so [F1] gives $f_{s^{\mathrm{\prime }}}(t,t)=2$ and $f_{s^{\mathrm{\prime }}}(1,t)=f_{s^{\mathrm{\prime }}}(t,1)=0$. Thus $f_{s^{\mathrm{\prime }}}=\delta u$ for the one-cochain with $u(t)=1$.

This makes the section-change formula visible in a concrete computation.

Because $M=C_2$ is written additively, the cocycle sum satisfies $(f+f)(t,t)=1+1=0$. So $f+f=0$ as a cocycle.

By [L1], the Baer sum of the extension class of $f$ with itself corresponds to the cohomology class of $f+f$, which step 1.1 identified with $0$. Thus the self-Baer-sum is the split class.

So this example realizes a nontrivial class whose double is zero.

The map $\pi$ is surjective, with kernel $\{0,\bar{3},\bar{6}\}$. Both $i_1$ and $i_2$ identify $C_3$ with that kernel, so $1\to C_3\stackrel{i_1}{\to }E\stackrel{\pi }{\to }{C}_3\to 1,1\to C_3\stackrel{i_2}{\to }E\stackrel{\pi }{\to }{C}_3\to 1$ are two extensions of $C_3$ by $C_3$ with the same middle group $E$.

Any automorphism of $E=C_9$ has the form ${\varphi }_u(\bar{x})=u\bar{x}$ with $u\in (\mathbb{Z}\mathrm{/}9\mathbb{Z}{)}^{\times }$. If an extension equivalence existed, it would satisfy $\varphi \circ i_1=i_2$ and $\pi \circ \varphi =\pi$. The first identity gives $3u\equiv 6(\mathrm{m}\mathrm{o}\mathrm{d}9)$, so $u\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}3)$. The second gives $u\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}3)$. This contradiction shows that no extension equivalence can exist.

Thus the same middle group supports two inequivalent extension structures, refuting [L1].