I came across an exercise in Alexander Beilinson's note refered by Tom here.
The exercise is:
Let $G$ be a finite group, M and N be $\mathbb{Z}[G]$-module that is free as $\mathbb{Z}$-module.
Show following are equivalent for $\mathbb{Z}[G]$-homomorphism $f:M\to N$.
(a) There is a $\mathbb{Z}$-homomorphism $h:M\to N$ and $f(x)=\sum_{g \in G}ghg^{-1}(x)$.
(b) There is a free $\mathbb{Z}[G]$-module $R$ and $f$ can factored by two $\mathbb{Z}[G]$-homomorphism $f_1:M \to R$ and $f_2:R \to N$.
I observed an example $G=\{g,g^2,g^3,g^4,g^5,g^6=1\}, M=N=\mathbb{Z}^2, g(a,b)=(a,-b)$, so $g^2m=m, h(a,b)=(a,b), f(a,b)=3(a,b)+3(a,-b)$ to construct below.
My solution:
Let $m_1,m_2,..$ be the generators of $M$ as $\mathbb{Z}$-module. Let $ H_k = \{ g|\ g(m_k)=m_k \}$.
Assume (a).
Take $r_1$ as the first generator of $R$ and define $f_1(m_1) = \sum_{g \in H_1} g(r_1)$, and $f_2(r_1)= |G/H_1| \cdot h(m_1)$. This satisfy (b) for $m_1$, and thus for any $m \in M$ generated by $m_1$, by compatibility of $\mathbb{Z}[G]$-module.
If there remains any generator $m_k$ not generated by $m_1$, repeat this as:
Take $r_k$ as a new generator of $R$ and define $f_1(m_k) = \sum_{g \in H_k} g(r_k)$, and $f_2(r_k) = |G/H_k| \cdot h(m_k)$.
Assume (b).
For $g \in H_1$, $f_1(m_1) = f_1(g(m_1)) = g(f_1(m_1))$. So $f_1(m_1)$ can be write as $\sum_{g \in H_1} g(r_1)$ by some $r_1 \in R$. Then take a representative coset $G/H_k$ and define $h(g(m_1))=f_2(g(r_1))$ if $g \in G/H_k$ and $h(g(m_1))=0$ otherwise. This satisfies (a).
If there remains any generator $m_k$ not generated by $m_1$, repeat this.
Question:
(1) I have a concern in my argument. For example if I take $m_1=(1,1),m_2=(0,1)$, the first step defines $f_1(m)$ for every $m = s(1,1)+t(1,-1)$ where $s,t \in \mathbb{Z}$. The second step is to define $f_1(m_2)$ but $f_2(2m_2)=f_2(0,2)$ have already defined in the first step, so it is not trivial whether I can make them compatible in such situation. Is there some way to fix this problem?
(2) I think I essentially decomposed $M$ into some irreducible modules. But I am not sure I can generally decompose $M = M_1 \oplus M_2 \oplus...$ such that each $M_k$ is free as $\mathbb{Z}$ module, especially of the form $M_k = \mathbb{Z}^{|G/H_k|}$ where $ H_k = \{ g|\ g(m_k)=m_k \}$. Is there some counterexamples?
[Edit] How about injecting $M$ into $M'=M\otimes \mathbb{Q}$ and decompose $M' = M'_1 \oplus M'_2 \oplus...$ such that $M'_k$ is irreducible and free as $\mathbb{Q}$ module and then take $M_k = M'_k \cap M$? (I don't have much experience with module over group)
(3) I appreciate any related information.
EDIT (observation after the answer)
I made some observation based on the answer. They gave good experience for me.
The condition (a) can be described as $M \to N$ can be factored into
$M \to \mathbb{Z}[G] \otimes M \to \mathbb{Z}[G] \otimes N \to N$,
$m \to \sum g \otimes g^{-1}(m)$, $1 \otimes m \to 1 \otimes h(m)$, $g \otimes n \to g(n) $.
Then (a)⇒(b) is obvious.
For the inverse implication, I noticed free $\mathbb{Z}[G]$-module $R$ is injective module and $M \to \mathbb{Z}[G] \otimes M$ is injective, so $M \to R$ factors through $\mathbb{Z}[G]\otimes M$. Similarly, $R$ is projective module and $\mathbb{Z}[G] \otimes N \to N$ is surjective so $R \to N$ factors through $\mathbb{Z}[G]\otimes N$.
It remains to show that I can factor them so that the $\mathbb{Z}[G] \otimes M \to \mathbb{Z}[G] \otimes N$ sends $1 \otimes m \to 1 \otimes h(m)$ for some $h$. I could comfirm this directly, and I also noticed this describes the bijection
$ {\mathrm{Hom}}_{\mathbb{Z}[G]} (\mathbb{Z}[G] \otimes M, N) \cong {\mathrm{Hom}}_{\mathbb{Z}} (M,N)$ obtained from forgetful and free adjoint functors between $\mathbb{Z}$-mod category and $\mathbb{Z}[G]$-mod category.
Here is a simple solution.
(a)$\implies$(b): Regard $\mathbb{Z}[G]\otimes_\mathbb{Z}N$ as a free $\mathbb{Z}[G]$-module by $$ g(x\otimes n)=(gx)\otimes n. $$ Define a $\mathbb{Z}[G]$-module homomorphism $f_1\colon M\to \mathbb{Z}[G]\otimes_{\mathbb{Z}}N$ by $$ f_1(m)=\sum_{g\in G}g\otimes h(g^{-1}m), $$ and $f_2\colon \mathbb{Z}[G]\otimes_{\mathbb{Z}}N \to N$ by $$ f_2\biggl(\sum_{g\in G}g\otimes n_g\biggr) = \sum_{g\in G}gn_g. $$ Then we have $f_2\circ f_1 = f$, so (b) holds.
(b)$\implies$(a): We can write $R\simeq \mathbb{Z}[G]\otimes_\mathbb{Z}L$ where $L$ is a free $\mathbb{Z}$-module. Define $\mathbb{Z}$-module homomorphisms $h_g\colon M\to L\;(g\in G)$ by $$ f_1(m)=\sum_{g\in G}g\otimes h_{g^{-1}}(m). $$ Since $f_1$ is a $\mathbb{Z}[G]$-module homomorphism, we have $h_g(m)=h_e(gm)$. Therefore we can write $$ f(m)=\sum_{g\in G}gf_2(h_e(g^{-1}m)), $$ so (a) is satisfied for $h=f_2\circ h_e$.