Let $p$ be an odd prime. Let $\zeta_p=e^{2\pi{\bf i}/p}$ and let $1\le k\le p-1$ be a divisor of $p-1$. Recently, when I learnt algebraic number theory, I met the following problem.

If we let 
$$U_k=\{y\in\mathbb{Z}/p\mathbb{Z}:\ y^k\equiv 1\pmod p\}$$
be the subgroup of all $k$-th root of unity over the finite field $\mathbb{Z}/p\mathbb{Z}$, then we can define an element 
$$x_k=\sum_{y\in U_k}\zeta_p^y.$$
Using the Galois theory, it is easy to see that $x_k$ is indeed contained in the unique subfield $M_k$ of $\mathbb{Q}(\zeta_p)$ with $[M_k:\mathbb{Q}]=(p-1)/k$.

 **I want to know if $x_k$ is exactly the generator of the subfield $M_k$, i.e., $\mathbb{Q}(x_k)=M_k$.** For example, when $k=1$ or $2$, the answer is positive.

Thank you for you comments!