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

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.

• What is $G$? Any group? And are these meant to be $\mathbb{Z}[G]$-modules? Commented Jan 23, 2018 at 12:37
• @TobiasKildetoft Yes and yes. Commented Jan 23, 2018 at 12:50
• You don't want $G$ to be finite? I'm not sure I believe this for infinite groups in general. Commented Jan 24, 2018 at 11:18
• @JeremyRickard I do not know how to prove the statement even in the case of a finite group and would be grateful if you could provide me with a hint/solution. The general statement may be wrong (well, I cannot be sure, since I know neither the proof nor counterexamples), but the original source (a professor of mine) stated the problem without the finiteness assumption. Commented Jan 24, 2018 at 11:34

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.
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.
• Nice. In fact, the same argument works if you take $M=\mathbb{F}_p$ with $G=\mathbb{Z}$ acting trivially. Commented Jan 25, 2018 at 6:46
• @JeremyRickard Yes, thank you. Also, we can take any non-free abelian group with $G=\mathbb{Z}$ acting trivially. An exotic counterexample could be $M=\underset{i\in\mathbb{N}}{\prod}\mathbb{Z}$. Commented Jan 25, 2018 at 9:19