 Solution

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