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Be $\mathbb{L}$ the splitting field, and be $G$ the Galois Group of $\mathbb{L}/\mathbb{Q}$,

I've to prove that $G \cong A \ltimes B $ with $|A| = 11$ (cyclic) and $|B|=10$, abelian.

The central idea should be considering intermediate fields like $\mathbb{Q}(\zeta_{11})$ and using lemmas cited in Kaplansky's "Fields and Rings" book in page 33 and 34, but I'm not very confident with semidirect products and I need some help...

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  • $\begingroup$ Can you explicitly write some generators of $\mathbb L$ over $\mathbb Q$? Hint: there are two natural generators which have degrees $11$ and $10$ over $\mathbb Q$. $\endgroup$ – user59193 Jul 18 '15 at 13:51
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Step 1 : prove that $\mathbb{L}=\mathbb{Q}(\zeta_{11}, \sqrt[11]{7})$

Step 2: Consider the extension $\mathbb{Q}(\zeta_{11})$. Is a Galois Extension? What about $Gal (\mathbb{Q}(\zeta_{11})/\mathbb{Q})$?

Step 3: Consider the extension $\mathbb{Q}(\sqrt[11]{7})$. Is a Galois Extension? What about $Gal (\sqrt[11]{7})/\mathbb{Q})$?

Step 4: What can you say about $\mathbb{Q}(\sqrt[11]{7}) \cap \mathbb{Q}(\zeta_{11})$? What meaning has if you tink to it form the Galois Correspondence point of view?

Finally I recall the following usefull result:

Theorem Let $G$ me a finite group with $H< G$ normal subgroup. If there exist $K<G$ suh that $KH=G$ and $H\cap K = \{e\}$, then $G$ is a semidirect product of $K$ and $H$.

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  • $\begingroup$ What you wrote in steps 1 to 3 is clear, but as I said I have some difficulties in proving that one of the 2 groups is normal and that $KH$ really gives $G$ $\endgroup$ – Klest Dedja Jul 18 '15 at 17:02
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    $\begingroup$ Normality: let $\zeta_0$ be any eleventh root of unity unequal to $1$, and thus primitive. Then any field automorphism at all sends $\zeta_0$ to a power of itself, and the extension $\Bbb Q(\zeta_0)\supset\Bbb Q$ is normal. $\endgroup$ – Lubin Jul 18 '15 at 19:09
  • $\begingroup$ To prove that $KH=G$ look simply at the cardinality of the product and use the fact that $\mathbb{Q}(\sqrt[11]{7}) \cap \mathbb{Q}(\zeta_{11})= \emptyset$ implies that $K \cap H =\{e\}$ $\endgroup$ – Sabino Di Trani Aug 8 '15 at 15:14

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