# Is it true that $\oplus_G H \cong \mathbb{Z}[G] \otimes_{\mathbb{Z}} H$ as $\mathbb{Z}[G]$ module?

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?

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