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Let $d>1$ be an odd integer. Given a simplicial set $X$ and $[\gamma]\in H^2(X,\mathbb{Z}/d)$, there exists a fibration $N\mathbb{Z}/d\to E\to X$, with $$E= X_\gamma:=N\mathbb{Z}/d\times_{\gamma} X.$$ Conversely, for any "central extension" $N\mathbb{Z}/d\to E\to X$, we have that $E\cong X_\gamma$, for some normalized cocycle $\gamma:X_2\to \mathbb{Z}/d$. I am interested in simplicial sets $X,E$ with a certain abelian-group-like structure and I would like to understand when $N\mathbb{Z}/d\to E\to X$ splits, assuming that we know explicitly $H^2(X,\mathbb{Z}/p)$.

For that purpose, a first step would be to understand the following particular case: let us assume that $d=p$ an odd prime and $X=(\mathbb{Z}/p)^n$. If a central extension $$\mathbb{Z}/p\to E\to (\mathbb{Z}/p)^n$$ satisfies that $E$ is abelian and $g^p=1$, for every $g\in E$, then $[\gamma]=0$. This is true by the classification of finite abelian groups, but this kind of classification theorem does not work more generally.

Is it possible to give a "cohomological" proof of the vanishing of $[\gamma]$? It means that we can assume $H^{\ast}((\mathbb{Z}/p)^n,\mathbb{F}_p)\cong \Lambda(x_1,\dots,x_n)\otimes \mathbb{F}_p(y_1,\dots.y_n)$ is known.

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    $\begingroup$ Yes. The terms in the exterior algebra determine the commutators of lifts of elements of the quotient group, while the terms in the polynomial algebra determine the $p$th powers. This can be see by restricting to two generator, respectively one generator subgroups. $\endgroup$ Commented Jan 23 at 11:29
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    $\begingroup$ @DaveBenson Thank you! It took me a while to see what you mean: for instance, $H^2(\mathbb{Z}/p\times\mathbb{Z}/p,\mathbb{F}_p)\cong \mathbb{F}_p\{x_1x_2,y_1,y_2\}$, where the extension induced by $x_1x_2$ is not abelian, while those induced by $y_1,y_2$ are abelian but their correspondig extension are not $p$-torsion but $p^2$-torsion. $\endgroup$
    – Antoine
    Commented Jan 30 at 16:09

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