A <crossed homomorphism> is a map $\phi:G\to M$ satisfying
$$
\phi(gh)=\phi(g)+g\phi(h).
$$
For a one-cochain, the formula in part i gives
$$
(d\phi)(g,h)=g\phi(h)-\phi(gh)+\phi(g),
$$
so 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
$$
\boxed{H^1(G,M)=\frac{\{\text{crossed homomorphisms }G\to M\}}
{\{g\mapsto gm-m:m\in M\}}}.
$$

Solution

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

Write every element as $a^kb^l$ 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 $a$ shows
$$
\phi(xy)=\phi(x)+x\phi(y),
$$
so $\phi$ is a <crossed homomorphism> and hence a <one-cocycle>.

It cannot be principal: if $\phi(g)=gm-m$ for some $m\in\mathbb Z$, then at $g=a$ one would have
$$
1=\phi(a)=a m-m=-2m,
$$
which is impossible in the <integers>. Thus $[\phi]\ne0$ and
$$
\boxed{H^1(D_n,M)\ne0}.
$$

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

{wiki}

A crossed homomorphism from $G$ to a $G$-module $M$ is a map $f:G\to M$ satisfying
$$
f(gh)=f(g)+g\cdot f(h).
$$
Thus crossed homomorphisms are precisely <one-cocycles>.

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 Incoming links: Crossed homomorphism