When $n \ge 4$, the third homology group $H_3(S_n,\mathbb{Z})$ of the symmetric group $S_n$ contains $\mathbb{Z}_{12}$ as a summand. Using the universal coefficient theorem we get $\mathbb{Z}_{12}$ as a summand of the cohomology group $H^3(S_n,\mathbb{Z}_{12})$.

**What's a nice 'formula' for a $\mathbb{Z}_{12}$-valued 3-cocycle on $S_n$ that generates this summand of the third cohomology?**

In more concrete terms, I'm looking for a nontrivial recipe to get an element $c(g,g',g'') \in \mathbb{Z}_{12}$ from three elements of $S_n$, such that

$$ c(g',g'',g''') - c(gg',g'',g''') + c(g,g'g'',g''') - c(g,g',g''g''') + c(g,g',g'') = 0 $$

for all $g,g',g'',g''' \in S_n$. Here "nontrivial" means that for all nonzero $\alpha \in \mathbb{Z}_{12}$, we do not have

$$ \alpha c(g,g',g'') = f(g',g'') - f(gg',g'') + f(g,g'g'') - f(g,g') $$

for some $f \colon S_n \times S_n \to \mathbb{Z}_{12}$.