Is $\text{Hom}_{\mathbb Z[G]}(M,\mathbb Z[G])$ finitely generated?

Question

Let $G$ be a finite group and $M$ a finitely generated $\mathbb Z[G]$-module. Is $\text{Hom}_{\mathbb Z[G]}(M,\mathbb Z[G])$ also finitely generated?

I think the answer is yes, and here is my (failed) attempt. Could you help me completing the argument, or providing a more insightful proof?

If $M$ is f.g. then there exist an $r\geq0$ and a surjective map $\mathbb Z[G]^r\to M$. Taking Hom we obtain that $\text{Hom}_{\mathbb Z[G]}(M,\mathbb Z[G])\to\text{Hom}_{\mathbb Z[G]}(\mathbb Z[G]^r,\mathbb Z[G])$ is injective, and $$\text{Hom}_{\mathbb Z[G]}(\mathbb Z[G]^r,\mathbb Z[G])\simeq \mathbb Z[G]^r$$ so $\text{Hom}_{\mathbb Z[G]}(M,\mathbb Z[G])$ is a submodule of $\mathbb Z[G]^r$. Thus it suffices to show that the submodules of $\mathbb Z[G]^r$ are f.g. But I've failed to prove that so far.

Answer 1

$\mathbb Z[G]^r$ is a finitely generated abelian group because $G$ is finite; therefore so are all its subgroups. Moreover, finitely generated as an abelian group automatically implies finitely generated as a $G$-module.

You could see this as expressing that $\mathbb Z[G]$ is a left noetherian ring.

Answer 2

This follows from the fact that $ℤ[G]$ is noetherian as a left-module over itself.

As a $ℤ$-module, $ℤ[G] \cong \bigoplus_{g ∈ G} ℤg$ is finitely generated, hence it is noetherian $ℤ$-module. Hence $ℤ[G]$ is also noetherian as $ℤ[G]$-module, and thus so is $ℤ[G]^r$. Submodules of noetherian modules are finitely generated.