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A question from group homology:

Let A and B be G - modules. Why is their tensor product over the integers (which I denote A@B) naturally a G - module? I have read that the natural action of G on A@B is g(a@b)=ga@gb however, I am uncertain as to why (g+h)(a@b)=g(a@b)+h(a@b).

A question from representation theory:

Let k be a field of characteristic 0. Let A and B be k[G] - modules. Then is A tensor B (over k) naturally a k[G] - module with the action defined as in the first question.

Thanks for any replies and sorry for the lack of latex.

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 If you consider $g+h$ as an element in the group ring $\mathbb{Z}[G]$, then the relation in question follows right from the definition: $(\sum_{g \in G} a_g\cdot g) \cdot (a \otimes b) := \sum_{g \in G} a_g \cdot (ga \otimes gb)$ (one says the action of G is extended linearly). This should also answer the second question. – Ralph Mar 3 2012 at 18:26
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 Your formula g(a@b)=ga@gb is only valid for group elements g, not for their formal sums. If you want a formula for formal sums, you'll have to use the comultiplication of the coalgebra $k\left[G\right]$. – darij grinberg Mar 3 2012 at 18:30
Maybe, but there is certainly some confusion. – Geoff Robinson Mar 3 2012 at 19:28

closed as off topic by Qiaochu Yuan, Mark Grant, Angelo, Alain Valette, Andreas Thom Mar 3 2012 at 20:33

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