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...
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 $H$.