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Let$K$ be the smallest splitting field $X^5-7$ over $\mathbb{Q}$. Then how many intermediate fields of $K$ which is Galois expansion of $\mathbb{Q}$.

I know $[K:\mathbb{Q}]=[K:\mathbb{Q}(7^{\frac{1}{5}})][\mathbb{Q}(7^{\frac{1}{5}}):\mathbb{Q}]=20$ by translation theorem. I guess Galois group $Gal(K/\mathbb{Q})$ is $\mathbb{Z}/{5\mathbb{Z}} \rtimes \mathbb{Z}/{4\mathbb{Z}}$.But I can't prove and I don't know how to find subgroup of $\mathbb{Z}/{5\mathbb{Z}} \rtimes \mathbb{Z}/{4\mathbb{Z}}$.

J.smith
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1 Answers1

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The Galois group is indeed isomorphic to $$ \mathbb{Z}/5 \rtimes (\mathbb{Z}/5)^{\times}\cong \mathbb{Z}/5 \rtimes \mathbb{Z}/4, $$

see here:

Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$.

Also the subgroups of such semidirect products have been determined at this site, e.g., here:

Subgroups of the Semi-Direct Product $\mathbb{Z}/\mathbb{7Z} \rtimes (\mathbb{Z}/\mathbb{7Z})^{\times}$

Dietrich Burde
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