My question is about the following theorem:
A polynomial $f(x)$ can be solved by radicals if and only if its Galois group is solvable.
The difficulty arises in the last assertion in this paragraph:
Suppose first that $f(x)$ can be solved by radicals. Then each root of $f(x)$ is contained in an extension as in the lemma [each root is contained in a root extension $K$ that is Galois over $F$ and each extension $K_{i+1}/K_i$ is cyclic] The composite $L$ of such extensions is again of the same type. Let $G_i$ be the subgroups corresponding to the subfields of $K_i$. Since $Gal(K_{i+1}/K_i) = G_i/G_{i+1}$, it follows that the Galois group $G = Gal(L/F)$ is solvable.
My difficulty in understanding comes at the last assertion, that $G$ is solvable. For some reason I am unable to see why that should be true. Any help in clarifying this would be appreciated.