Take the 2-minute tour ×  I know that
$H^3(S_3,\mathbb{Z}_3)\cong \mathbb{Z}_3$ (S_3 is the symmetric
group for three elements). So this group is generated by any
nontrivial cocycle. But I don't know how to explicitly find such
a cocycle (any explicit representation will do - even a table
describing all possible values).  How can I find the cocycle in
this case, and is there a general way of doing it?
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$\newcommand\ZZ{\mathbb{Z}}$Using the Hochschild-Lyndon-Serre
spectral sequence for the extension $$1\to C_3\to S_3\to
C_2\to1$$ (here $C_n$ is a cyclic group of order $n$) and
checking that the action of $C_2$ on $H^3(C_3,\ZZ_3)$ is
trivial, we see that restriction induces an isomorphism
$$\mathrm{res}:H^3(S_3,\ZZ_3)\to H^3(C_3,\ZZ_3),$$ and, in
particular, by the usual computation of the cohomology of cyclic
groups, we see that $H^3(S_3,\ZZ_3)$ is cyclic of order $3$.
Moreover, the general theory tells us that the composition
$$H^3(S_3,\ZZ_3)\xrightarrow{\ma