A field extension of the form $\mathbb{Q}(\zeta_n, \sqrt[n]{\beta})$ where $\zeta_n$ is a primitive $n$th root of unity and $\beta \in \mathbb{Q}(\zeta)$ is called a Kummer extension.
Even though $\mathbb{Q}(\zeta_n, \sqrt[n]{\beta})/\mathbb{Q}(\zeta_n)$ and $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ are Galois and Abelian it is not always the case that $\mathbb{Q}(\zeta_n, \sqrt[n]{\beta})/\mathbb{Q}$ is Galois.
Question: If $\mathbb{Q}(\zeta_n, \sqrt[n]{\beta})/\mathbb{Q}$ is Galois will it be Abelian ?
No. It can be dihedral: take $n=3$ and $\beta=2$ to get the splitting field of $x^3-2$ as your field $\mathbb{Q}(\zeta_n, \sqrt[n]{\beta})$. The Galois group over $\mathbb{Q}$ is a transitive subgroup of $S_3$ (thought of as the dihedral group of order $6$) containing a transposition (complex conjugation) and therefore equal to $S_3$.