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I noticed the following funny fact when studying cohomology of finite groups. I explain it in the case of $H^2$ but it generalizes.

Consider a two-cocycle $a\in Z^2(G,A)$ where $A$ is a left G-module. Explicitly, this means that $$ g_1 a(g_2,g_3) = a(g_1g_2,g_3)-a(g_1,g_2g_3)+a(g_1,g_2). $$

I realized that if one considers a free module $\mathcal{A}$ generated by symbols $a_{g,h}$ (with trivial relations $a_{e,g}=a_{g,e}=0$), the same expression $$ g_1 a_{g_2,g_3} = a_{g_1g_2,g_3}-a_{g_1,g_2g_3}+a_{g_1,g_2}. $$ makes $\mathcal{A}$ a $G$-module. Tautologically, a cochain defined by $$ a(g,h) := a_{g,h} $$ is a cocycle in $Z^2(G,\mathcal{A})$, and any cocycle in $Z^2(G,A)$ comes from this tautological cocycle and a homomorphism $\mathcal{A}\to A$.

Is this $G$-module $\mathcal{A}$ some well-known object? I think it should be, but I do not recognize it.

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    $\begingroup$ Just a guess but it may just be the group ring $\Bbb Z[G]$ $\endgroup$
    – Ali Caglayan
    Commented Dec 27, 2017 at 12:39
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    $\begingroup$ This looks like dimension shifting. (Brown, Cohomology of groups, Chap. III.7, page 74.) $\endgroup$
    – Thibaut Dumont
    Commented Dec 27, 2017 at 13:20
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    $\begingroup$ You are observing that $Z^2(G,A) = \hom_G(\mathcal A,A)$, where $\mathcal A=\mathbb ZG\otimes \mathbb ZG$ is a $G$-module with your given left action. $\endgroup$
    – Pedro
    Commented Dec 27, 2017 at 18:28

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