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David White
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I don't know the algebraic side of cup products in group cohomology, but they can indeed be identified with topological cup products. The group cohomology H^n(G;A)$H^n(G;A)$ can be identified with the cohomology H^n(BG;A)$H^n(BG;A)$ of the classifying space BG$BG$ of G$G$ with coefficients in A (interpreting the action of G$G$, which is the fundamental group of BG$BG$, on A as making A a local coefficient system on BG$BG$). The cup product on group cohomology is then the same as the usual cup product of singular cohomology.

I don't know the algebraic side of cup products in group cohomology, but they can indeed be identified with topological cup products. The group cohomology H^n(G;A) can be identified with the cohomology H^n(BG;A) of the classifying space BG of G with coefficients in A (interpreting the action of G, which is the fundamental group of BG, on A as making A a local coefficient system on BG). The cup product on group cohomology is then the same as the usual cup product of singular cohomology.

I don't know the algebraic side of cup products in group cohomology, but they can indeed be identified with topological cup products. The group cohomology $H^n(G;A)$ can be identified with the cohomology $H^n(BG;A)$ of the classifying space $BG$ of $G$ with coefficients in A (interpreting the action of $G$, which is the fundamental group of $BG$, on A as making A a local coefficient system on $BG$). The cup product on group cohomology is then the same as the usual cup product of singular cohomology.

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Eric Wofsey
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I don't know the algebraic side of cup products in group cohomology, but they can indeed be identified with topological cup products. The group cohomology H^n(G;A) can be identified with the cohomology H^n(BG;A) of the classifying space BG of G with coefficients in A (interpreting the action of G, which is the fundamental group of BG, on A as making A a local coefficient system on BG). The cup product on group cohomology is then the same as the usual cup product of singular cohomology.