Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 3 / 16F / iii / Solution

Let $\alpha =\sqrt[5]{3}$ and let $\zeta$ be a primitive fifth root of unity. Then $L=\mathbb{Q}(\alpha ,\zeta )$. The field degrees $[\mathbb{Q}(\alpha ):\mathbb{Q}]=5$ and $[\mathbb{Q}(\zeta ):\mathbb{Q}]=4$ both divide $[L:\mathbb{Q}]$, while adjoining the two generators gives $[L:\mathbb{Q}]\le 20$. Consequently $[L:\mathbb{Q}]=20$.

The cyclotomic field $\mathbb{Q}(\zeta )/\mathbb{Q}$ is Galois with cyclic group of order $4$. The Fundamental theorem of Galois theory gives a normal subgroupThus $N\cong C_5$ and the chain $\boxed{G▹N▹\{1\}}$ has abelian successive quotients, proving that $G$ is a solvable group. One may refine it by inserting the inverse image of the order-two subgroup of $C_4$, giving cyclic factors $C_2,C_2,C_5$.