Let $G$ be a finite group and let $K$ be a field of characteristic $0$. Let $M$ be a free, finitely generated $K$-module and assume in addition that $M$ is cyclic as a (left) $K[G]$-module.
Consider the dual $M^{\ast}:=Hom_{K}(M, K)$ of $M$. I know that there is a non-canonical isomorphism of $K$-modules $M \cong M^{\ast}$ and that $M^{\ast}$ can be made into a $K[G]$-module.
Can we prove something about the structure of $M^{\ast}$ as a $K[G]$-module, for example can we prove that it is cyclic, since $M$ is cyclic?
Yes, it’s cyclic.
A cyclic module $M$ is just one that is a quotient of $K[G]$, the regular $K[G]$-module. But by Maschke’s theorem, a quotient of $K[G]$ is a direct summand, so $M^*$ is a direct summand of $K[G]^*$, which is isomorphic to $K[G]$, so $M^*$ is also cyclic.
But in positive characteristic, where Maschke’s theorem doesn’t apply, it’s not always true that the dual of a cyclic module is cyclic.