Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 4

The inhomogeneous group cochain group iswith $C^0(G,M)=M$. Its group coboundary $d:C^r(G,M)\to C^{r+1}(G,M)$ isOne checks that $d^2=0$, and group cohomology is $H^r(G,M)=\ker d/\mathrm{im}d$.

A crossed homomorphism is a map $\varphi :G\to M$ satisfyingFor a one-cochain, the formula in part i givesso the crossed homomorphisms are exactly the one-cocycles. A zero-cochain $m\in M$ has coboundary $g\mapsto gm-m$, the principal crossed homomorphism associated with $m$. Therefore

For $g,h\in G$, substitute the definition of $\psi$ and in the first sum replace $\gamma$ by $\delta g$:The two-cocycle identity at $(\delta ,g,h)$ says that the expression in parentheses is $\delta \varphi (g,h)$. Every summand is therefore $\varphi (g,h)$, and

Let $\varphi \in Z^2(G,\mathbb{Q})$. Part iii gives $d\psi =|G|\varphi$. Since division by $|G|$ is possible in the rational numbers,is a two-coboundary. HenceThis is a degree-two instance of the vanishing of finite-group cohomology when the group order is invertible on the coefficient module.

For the trivial action, crossed homomorphisms $G\to \mathbb{Q}$ are ordinary group homomorphisms, while principal crossed homomorphisms vanish. The image of a finite group in the torsion-free additive group $\mathbb{Q}$ must be trivial, soDegree-zero cohomology is the invariant submodule; the action is trivial, so

The short exact sequence of $G$-modules induces the long exact sequence in group cohomology

Apply part i to the short exact sequence of trivial $G$-modulesThe relevant segment isBoth outer groups vanish by parts a(iv) and a(v), so the connecting homomorphism is an isomorphism:

The trivial action givesbecause $D_n$ is finite. By part b(ii),In the abelianization the relation $aba=b^{-1}$ becomes $b=b^{-1}$. HenceEach finite cyclic group is naturally isomorphic to its character group in $\mathbb{Q}/\mathbb{Z}$, so

Write every element as $a^k{b}^l$ with $k\in \{0,1\}$. The action of this element on $M=\mathbb{Z}$ is multiplication by $(-1{)}^k$. A direct check of the four possibilities for the two exponents of $a$ showsso $\varphi$ is a crossed homomorphism and hence a one-cocycle.

It cannot be principal: if $\varphi (g)=gm-m$ for some $m\in \mathbb{Z}$, then at $g=a$ one would havewhich is impossible in the integers. Thus $[\varphi ]\ne 0$ and