FALSE: a polynomial solvable by radicals must have abelian Galois group

Let $\alpha =\sqrt[5]{2}$, $E=\mathbb{Q}(\alpha )$, and $F=\mathbb{Q}({\zeta }_5)$. By [L2] with the prime $2$, $x^5-2$ is irreducible and $[E:\mathbb{Q}]=5$. By [L3], $F\mathrm{/}\mathbb{Q}$ is Galois of degree $4$. The degree of $E\cap F$ divides both $5$ and $4$ by [L5], so $E\cap F=\mathbb{Q}$. Hence [L4] gives $[K:\mathbb{Q}]=20$, $[K:F]=5$, and $[K:E]=4$.

The minimal polynomial of $\alpha$ over $F$ is therefore $x^5-2$. Since ${\zeta }_5\alpha$ is another root in $K$, [L6] gives an $F$-automorphism $\sigma$ of $K$ with $\sigma (\alpha )={\zeta }_5\alpha$ and $\sigma ({\zeta }_5)={\zeta }_5$; it has order $5$. Similarly, $[K:E]=4$ shows that the minimal polynomial of ${\zeta }_5$ over $E$ is the degree-four cyclotomic polynomial. Its root ${\zeta }_5^2$ therefore gives by [L6] an $E$-automorphism $\tau$ with $\tau (\alpha )=\alpha$ and $\tau ({\zeta }_5)={\zeta }_5^2$; it has order $4$.

These automorphisms do not commute: $\tau \sigma {\tau }^{-1}(\alpha )=\tau ({\zeta }_5\alpha )={\zeta }_5^2\alpha ={\sigma }^2(\alpha )\mathrm{\ne }\sigma (\alpha )\mathrm{.}$ So the Galois group of $x^5-2$ is nonabelian, even though [L1] makes the polynomial solvable by radicals. This disproves the claim.