Are finite-dimensional representations of groups of type $\text{FP}_{\infty}$?

Question

Let $G$ be a group (possibly infinite) and $k$ be a field. A module $M$ over $k[G]$ is said to be of type $\text{FP}_{\infty}(k)$ if it has a projective resolution each of whose terms is finitely generated. We say that $G$ itself is of type $\text{FP}_{\infty}(k)$ if the trivial $k[G]$-module $k$ is of type $\text{FP}_{\infty}(k)$.

Assume that $G$ is a group of type $\text{FP}_{\infty}(k)$ and that $M$ is a $k[G]$-module that is finite-dimensional over $k$ (in other words, $M$ is a finite-dimensional representation of $G$). Must $M$ be of type $\text{FP}_{\infty}(k)$? If not, are there stronger finiteness properties that we can put on $G$ to assure that this holds (for instance, having a compact $K(G,1)$)?

Answer 1

Cleaned up answer (6/11/18) after comments of Andy Putman.

The answer is yes.

Thm 2 of https://www.tandfonline.com/doi/abs/10.1080/00927870600796110 shows that if G is $FP_\infty$ over $k$, then $kG$ has a free resolution as a bimodule by finitely generated free bimodules in each dimension.

If you tensor this resolution with $M$ over $kG$ you get a free resolution of $M$ with the finiteness properties you want. Tensoring with $M$ gives a resolution because its homology is $Tor^{kG}(M,kG)$.

It is easy to check that $(kG\otimes_k kG)\otimes_{kG} M\cong kG^{\dim M}$ as a left $kG$-module so that the free resolution is finitely generated in each degree. The basis as a $kG$-module is the tensors $1\otimes 1\otimes b$ with $b$ running over a basis of $M$.