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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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What I did at the end: A 3-cocyle is defined in this case as a function $f(a,b,c)$ satisfying $f(a,b,c)+f(a,bc,d)-f(b,c,d)-f(a,b,cd) = 0$ for all $a,b,c,d\in G$.

But this is a linear equation over $\mathbb{Z}_3$ with the variables being $f(a,b,c)$ for $a,b,c$, and can be solved easily using a computer.

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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 $$H^3(S_3,\ZZ_3)\xrightarrow{\mathrm{res}} H^3(C_3,\ZZ_3)\xrightarrow{\mathrm{cores}} H^3(S_2,\ZZ_3)$$ of the restriction map with the correstriction map is multiplication by the index of $C_3$ in $S_3$, namely $2$: this composition is then an isomorphism too. It follows immediately that to construct a generator of $H^3(S_3,\ZZ_3)$ it is enough to construct a cocycle whose class generates of $H^3(C_3,\ZZ_3)$, which is easy, and then corestrict it to a cocycle on $S_3$, which will generate $H^3(S_3,\ZZ_3)$. To do the latter, there are explicit formulas in textbooks (which I never remember)

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