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 this carry function with the extension class. If it were a coboundary, then [L2] would make the extension split, but $C_{p^2}$ has no subgroup of order $p$ complementary to its unique subgroup of order $p$.

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 section $g\mapsto (0,g)$ is then a homomorphism, so the extension splits. By [L1], this is precisely the zero class in $H^2(G,M)$.

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 kernel $A$. That kernel is central, so the extension is determined by the parameter $a:={\tilde{x}}^n$.

Replacing $\tilde{x}$ by $b\tilde{x}$ with $b\in A$ changes the parameter to $(b\tilde{x}{)}^n=nb+{\tilde{x}}^n$ because $A$ is central and written additively. Thus two parameters define equivalent extensions exactly when they differ by an element of $nA$.

So the central extension classes are parametrized by $A\mathrm{/}nA$, which is the familiar description of $H^2(C_n,A)$ from [L1].

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].