Show that $ A_G \simeq \mathbb{Z} \otimes_{\mathbb{Z}[G]}A $ for $G$-modules. I found this ring theory exercise in an algebra textbook.  Let $G$ be a group and then $\mathbb{Z}[G]$ is a module and let $A$ be a $G$-module.
$$ A_G \simeq \mathbb{Z} \otimes_{\mathbb{Z}[G]}A $$
Here the group of

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*$G$-invariants is: $ A^G = \{ a \in A  : g \cdot A = a \text{ for all } g\in G, a \in A\} $

*$G$-coinvariants is $A_G = A/I_G A$ the part of $A$ that is fixed by $G$
The Wikipedia article on tensor product has a lot of discussion but also some warnings, so there is also the article on Stacks Project.
 A: Credit to Shivering Soldier for pointing out the cleaner argument given below.

Lemma: For any ring $R$, ideal $I\leqslant R$, and left $R$-module $M$, we have $M\big/IM\cong R/I\otimes_R M$.
Proof: Define a map $\phi:M\rightarrow R/I\otimes_RM$ by $m\mapsto (1+I)\otimes m$. That $\phi$ is a morphism of abelian groups is immediate from the definition of the tensor product; indeed, for $m,n\in M$, we have $$\phi(m-n)=(1+I)\otimes(m-n)=(1+I)\otimes m-(1+I)\otimes n,$$ as needed. Furthermore, for any $i\in I$ and $m\in M$, we have $$\phi(i\cdot m)=(1+I)\otimes i\cdot m=(1+I)\cdot i\otimes m=(i+I)\otimes m=(0+I)\otimes m=0,$$ so $IM\subseteq\ker{\phi}$ and hence $\phi$ descends to a map $\overline\phi:M\big/IM\rightarrow R/I\otimes_R M$.
We now define $\overline\psi:R/I\otimes_R M\rightarrow M\big/IM$, an inverse to $\overline\phi$, via the universal property of tensor products. First define $\psi:R/I\times M\rightarrow M\big/IM$ by $(r+I,m)\mapsto rm+IM$. To see that this is well-defined, note that $\psi(i+I,m)=im+IM=0+IM$ for any $i\in I$, as needed. $\psi$ is clearly additive in each variable, and we claim that it is $R$-balanced. Indeed, for $x\in R$, $r+I\in R\big/I$, and $m\in M$, we have $$\psi((r+I)\cdot x,m)=\psi(rx+I,m)=rxm+IM=\psi(r+I,x\cdot m),$$ as desired.
Thus $\psi$ descends to a group homomorphism $\overline\psi:R/I\otimes_R M\rightarrow M\big/IM$. We claim now that $\overline\psi$ and $\overline\phi$ are mutual inverses. To see that $\overline\psi\circ\overline\phi=\operatorname{id}_{M/IM}$, let $m\in M$. Then $$\overline\psi(\overline\phi(m+IM))=\overline\psi((1+I)\otimes m)=1m+IM=m+IM,$$ as desired. To see that $\overline\phi\circ\overline\psi=\operatorname{id}_{R/I\otimes M}$, it suffices to show that it acts as the identity on simple tensors. Thus let $r+I\in R\big/I$ and $m\in M$. We have $\overline\psi((r+I)\otimes m)=rm+IM$, and $$\overline\phi(rm+IM)=(1+I)\otimes r\cdot m=(1+I)\cdot r\otimes m=(r+I)\otimes m,$$ again as desired. Hence $\overline\phi$ and $\overline\psi$ are mutual inverses and so we have the desired isomorphism.

The result you desire is now an immediate corollary of the above lemma. Just to make all notation clear: $\mathbb{Z}\otimes_{\mathbb{Z}[G]} A$ is an abelian group, where $\mathbb{Z}$ is considered as a (right) $\mathbb{Z}[G]$-module on which every $g\in G$ acts as the identity. We can define $\iota:\mathbb{Z}[G]\rightarrow \mathbb{Z}$, the "augmentation map", by extending $g\mapsto 1$ linearly for each $g\in G$. Then $I_G$, the "augmentation ideal", is defined to be $\ker{\iota}$, and $A_G:=A\big/I_GA$.
Note that $\iota$ is in fact surjective (for instance, $\iota(n1_G)=n$ for any $n\in\mathbb{N}$, where $1_G\in G$ is the identity element). Thus, by the first isomorphism theorem, we have $\mathbb{Z}\cong\mathbb{Z}[G]\big/I_G$. This induces the chain of isomorphisms $$\mathbb{Z}\otimes_{\mathbb{Z}[G]} A\cong\mathbb{Z}[G]/I_G\otimes_{\mathbb{Z}[G]}A\cong A\big/I_GA,$$ where the second isomorphism follows from the lemma above, and this gives the desired result.
A: $\Bbb{Z}$ is a $\Bbb{Z}[G]$ module through $g.n=n$
$A\to \mathbb{Z} \otimes_{\mathbb{Z}[G]}A$ is surjective
$1\otimes g.a=g.1\otimes a=1\otimes a$ so
$$\mathbb{Z} \otimes_{\mathbb{Z}[G]}A\cong \mathbb{Z} \otimes_{\mathbb{Z}[G]}(A/I_GA)$$
where $I_G=\{ \sum n_g g\in \Bbb{Z}[G], \sum n_g=0\}$ is the two-sided ideal generated by the $1-g$.
Clearly since $g$ acts trivially on $A/I_GA$ $$\mathbb{Z} \otimes_{\mathbb{Z}[G]}(A/I_GA)\cong A/I_GA$$
