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There is a theorem that says that if $E/F$ is a normal and radical extension with $\text{char}(K)=0$ then then $\text{Aut}(E/F)$ is soluble. But why do we need normal and $\text{char}(K)=0?$ What would happen if $E/F$ is not normal? For example $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not normal but its Galois group ${Id}$ is solvable? What would be a counter-example if $\text{char}(K) \neq 0$?

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  • $\begingroup$ Perhaps you mean solvable in the title? $\endgroup$ – Yanko Dec 28 '18 at 15:37
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    $\begingroup$ @Yanko solvable = soluble (British vs American English, don't know which one is which) $\endgroup$ – Kenny Lau Dec 28 '18 at 15:47
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See e.g. here for a proof of the more general theorem that requires neither normality nor zero characteristic.

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