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# We prove that F K1K2 also has a solvable Galois group.

Please help with some explanation of steps used

Suppose that F ⊂ K1 and F ⊂ K2 are Galois extensions, with solvable Galois groups. We prove that F ⊂ K1K2 also has a solvable Galois group.

(a) Show that restriction gives a homomorphism Gal(K1K2/K1) → Gal(K2/F), and show that this homomorphism is injective (HINT: look at generators of K2 over F and see if they generate K1K2 over K1).

(b) Argue that Gal(K1/F) Gal(K1K2/F) / Gal(K1K2/K1) (look up and refer to theorems you use!).

(c) Show why the above steps allow us to conclude that Gal(K1K2/F) is solvable.