Find the splitting field of $X^5-11$ over $\mathbb{Q}$ and describe the Galois group of this splitting field over $\mathbb{Q}$ The problem

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*Find $K$ the splitting find of $x^5 - 11$ over $\mathbb{Q}$


*Describe $\mathrm{Aut}(K/\mathbb{Q})$.
My works

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*The roots of $x^5 - 11$ over $\mathbb{Q}$ are $\sqrt[5]{11},\sqrt[5]{11}\omega,\sqrt[5]{11}\omega^2,\sqrt[5]{11}\omega^3,\sqrt[5]{11}\omega^4$. Then

$$ K = \mathbb{Q}(\sqrt[5]{11},\omega) $$


*We have $\vert \mathrm{Aut}(K/\mathbb{Q}) \vert = 20$ since

$$ [ K : \mathbb{Q} ] = [ K : \mathbb{Q}(\sqrt[5]{11}) ][ \mathbb{Q}(\sqrt[5]{11}) : \mathbb{Q} ] = 4 \cdot 5 = 20 $$
Indeed $[ \mathbb{Q}(\sqrt[5]{11}) : \mathbb{Q} ] = 5$ because $x^5 - 11$ is the minimal polynomial of $\sqrt[5]{11}$ over $\mathbb{Q}$. Then $[ K : \mathbb{Q}(\sqrt[5]{11}) ] = 4$ because $x^4 + x^3 + x^2 + x + 1$ is the minimal polynomial of $\omega$ over $\mathbb{Q}(\sqrt[5]{11})$
Before describing element by element the Galois group, I would like to be sure about the number of elements inside. Is my proof correct?
I am not sure about 2., did I write the right justifications in 1. and 2.?
Thank you
 A: I believe you are correct that $|{\rm Aut}(K : \mathbb Q)| = 20$.
Minor point: How do you know that $x^4 + x^3 + x^2 + x + 1$ is irreducible over $\mathbb Q(\sqrt[5]{11})$? I mean, it looks that way for sure, but can we prove it? Well, suppose for contradiction that $x^4 + x^3 + x^2 + x + 1$ is reducible over $\mathbb Q(\sqrt[5]{11})$ and the minimal polynomial of $\omega$ over $\mathbb Q(\sqrt[5]{11})$ is in fact a polynomial of degree $d$, where $d$ is either $1$, $2$ or $3$. Then $[K : \mathbb Q ] = 5d$. But $K$ contains the subfield $\mathbb Q(\omega)$, and we know that $[\mathbb Q(\omega) : \mathbb Q] = 4$. And $4$ does not divide $5d$ if $d$ is $1$, $2$ or $3$.
As for describing the structure of the ${\rm Aut}(K : \mathbb Q)$, I suggest we present the group as a semi-direct product.

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*Let $N$ be the subgroup of ${\rm Aut}(K : \mathbb Q)$ that fixes $\mathbb Q(\omega)$. $N$ must be normal (since $\mathbb Q(\omega)$ is normal over $\mathbb Q$); $N$ must be the order-five cyclic group generated by the automorphism $\sigma$ that sends $\omega \mapsto \omega$ and $\sqrt[5]{11} \mapsto \omega\sqrt[5]{11}$ (since each automorphism is determined by how it acts on the elements $\omega$ and $\sqrt[5]{11}$ that generate $K$ over $\mathbb Q$, and since each automorphism maps roots of $X^5 - 11$ to roots of $X^5 - 11$).


*Let $H$ be the subgroup of ${\rm Aut}(K : \mathbb Q)$ that fixes $\mathbb Q(\sqrt[5]{11})$. $H$ must be the order-four cyclic group generated by the automorphism $\psi$ that sends $\omega \mapsto \omega^2$ and $\sqrt[5]{11} \mapsto \sqrt[5]{11}$ (since the exponent $2$ is a primitive root in the group of units modulo $5$).
Since $N \cap H$ is the trivial group and since $|NH| = 20 = |{\rm Aut}(K : \mathbb Q)|$, it must be the case that ${\rm Aut}(K : \mathbb Q)$ is a semi-direct product of $N$ with $H$.
Now all that remains is for you to work out what the commutator $\psi \circ \sigma\circ \psi^{-1}$ is, so that you can describe the structure of this semi-direct product...
