A $G$-module admits a surjection from a $G$-module, which is free as an abelian group, such that the kernel is free

Question

I am asking for a hint on how to prove the following result:

Let $G$ be a group, $M$ a $G$-module (here “$G$-module” means $\mathbb{Z}[G]$-module). Show that there exists an exact sequence of $G$-modules $0\to K\to\tilde{M}\to M\to 0$ such that $K$ is a free $G$-module and $\tilde{M}$ is free as an abelian group.

I understand how to prove the result when the roles of $K$ and $\tilde{M}$ are interchanged. From the associated long exact sequence, one sees that the higher cohomology groups of $M$ and $\tilde{M}$ are isomorphic. However, it is not clear to me how to construct such a module: the only natural way I know to construct a $G$-module which is free as an abelian group yields a projective module.

Answer 1

Here's a hint/sketch for finite $G$.

Take a short exact sequence $$0\to N\to F\to M\to0$$ with $F$ a free module. Find a $\mathbb{Z}$-split inclusion $N\to K$ from $N$ into a free $\mathbb{Z}[G]$-module, and take the pushout.

Answer 2

The statement is false in general. Let $G=\mathbb{Z}$ and $M=\mathbb{F}_{p^n}$, where $n>1$ and $x$ acts as the Frobenius. The group ring is isomorphic to $\mathbb{Z}[x,x^{-1}]$. It is easy to see that the group of invariants of a free $\mathbb{Z}[x,x^{-1}]$-module is trivial. If there was a desired exact sequence, then one would have $\tilde{M}^G=M^G$ (it follows from the long exact sequence), which is impossible, because $M^G=\mathbb{F}_p$ and $\tilde{M}^G$ must be $\mathbb{Z}$-free.