TITLE: Is a $k[G]$-module which is free on every cyclic subgroup free?
QUESTION [7 upvotes]: Let $G$ be a finite group and let $M$ be a representation of $G$ over a field $k$. Suppose that, for every cyclic subgroup $C$ of $G$, we have $M|_{C} \cong k[C]^{\oplus [G:C]}$. Can we conclude that $M \cong k[G]$?
In characteristic zero, this is immediately true by character theory, but I don't know what happens in finite characteristic.
Motivation: I was thinking about the normal basis theorem in Galois theory, and I remembered that many sources do the cyclic case separately. Suppose that $M/k$ is a Galois extension with Galois group $G$. Then, for each cyclic subgroup $C$, we know that $M/\text{Fix}(C)$ is a Galois extension with Galois group $C$ so, if we knew the normal basis theorem for cyclic extensions, we would know that $M \cong \text{Fix}(C)[C] \cong k[C]^{\oplus [G:C]}$ as a $k[C]$-module. I was wondering whether this might already be enough to finish the proof with no more field theory input.

REPLY [10 votes]: Suppose that $E = C_p \times \dots \times C_p$ be an elementary abelian $p$-group, say of rank $r$, and let $k$ be a field of characteristic $p$. Then the group algebra $kE$ is isomorphic to $k[x_1, \dots, x_r]/(x_1^p, \dots, x_r^p)$. If $v$ is in the vector space spanned by the $x_i$, then $(1+v)^p=1$, so $1+v$ generates a subgroup of the group of units of $kE$. This is called a cyclic shifted subgroup of $kE$.
A result of Dade originally, then independently later proved by Carlson and Avrunin-Scott, says the following: A $kE$-module $M$ is free if and only if the restriction of $M$ to each cyclic shifted subgroup is free.
(Dade: Lemma 11.8 in Endo-Permutation Modules over p-Groups II, https://www.jstor.org/stable/1971169. Carlson: Theorem 4.4 in The varieties and the cohomology ring of a module, https://www.sciencedirect.com/science/article/pii/0021869383901217.)
I believe that there are examples of modules which restrict to free modules over every honest cyclic subgroup but not over every cyclic shifted subgroup, but I don't have one at my fingertips. There should be an example with $G = C_2 \times C_2$, I think.