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Let $H$ be an abelian group and $G$ be any group. Then is it true that $ \oplus_{G} H $ is isomorphic to $\mathbb{Z}[G] \otimes_{\mathbb{Z}} H$ as $\mathbb{Z}[G]$ module. I am asking it following the question Homology of $X_{\infty}$ space for a given Seifert surface of an oriented link $L$. asked me. In that question $H_i(Y' \cap Y''; \mathbb{Z})= \mathbb{Z}[t^{-1},t] \otimes_{\mathbb{Z}} H_i(F;Z)$ for $i=0,1$. But basic calculations suggest that $H_i(Y' \cap Y''; \mathbb{Z})= \oplus_{\langle t \rangle} H_i(F;\mathbb{Z})$, where $\langle t\rangle$ is isomorphic to $\mathbb{Z}$. Thus I am thinking that $\oplus_G H \cong \mathbb{Z}[G] \otimes_{\mathbb{Z}} H$ as $\mathbb{Z}[G]$ module may be true, but I am not able to define map or produce a counterexample.

Can anyone help me in resolving it?

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Define $\oplus_G H$ as ${\Bbb Z}[G]$-module:

$g' \cdot h_g = h_{g'g}$

and ${\Bbb Z}[G]\otimes_{\Bbb Z} H$ as ${\Bbb Z}[G]$-module:

$g'(g\otimes h) = (g'g)\otimes h$.

Then the morphism is given by

$h_g \mapsto g\otimes h$

with $g'\cdot h_g = h_{g'g}\mapsto (g'g)\otimes h = g'(g\otimes h)$.

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