I am trying to follow a proof of the Kronecker-Weber theorem (if $K$ is a number field with abelian Galois group then it is contained in a cyclotomic field) but I cannot understand one step.
I could understand that one can only consider the case where $K$ is an abelian extension of prime power degree (say $p^n$) such that $p$ is the only ramified prime. Specifically, one proves the following:
Every abelian extension of $\mathbb{Q}$ is the compositum of abelian extensions of prime power degree.
Let $G=\text{Gal}\left (K/\mathbb{Q}\right )$. Since $G$ is abelian, it is the direct sum of its Sylow subgroups. If $\left |G\right |=p_1^{n_1}\cdots p_r^{n_r}$ is the prime decomposition of the order of $G$, then we take $G_i$ to be the direct sum of all $p_j$-Sylow subgroups of $G$ with $j\neq i$. The fixed field $K_{G_i}$ has degree $p_i^{n_i}$ for each $i$, therefore $K$ is the composite of all $K_{G_i}$ because that composite has degree $p_1^{n_1}\cdots p_r^{n_r}$ (this is because the degrees are pairwise coprime). $\blacksquare$
Therefore the theorem is reduced to the case where $K$ is an abelian extension of prime power degree. Now, one can prove the following:
Let $K$ be an abelian extension of $\mathbb{Q}$ with $\left [K:\mathbb{Q}\right ]=p^m$. Suppose $q$ is a prime $\neq p$ which is ramified in $K$.
Fix $Q$ a prime of $K$ lying over $q$. Then the ramification index $e=e(Q|q)$ divides $q-1$. Let $L$ be the unique subgroup of $\mathbb{Q}(\xi)$ of degree $e$, where $\xi=e^{\frac{2\pi i}{q}}$.
Let $U$ be a prime of $KL$ lying over $Q$ and $K'$ denote the inertia field $\left (KL\right )_{E\left (U|q\right )}$. Then $E\left (U|q\right )$ is cyclic and $e(U|q)=e$. Also, $q$ is unramified in $K'$ and every prime of $\mathbb{Z}$ which is unramified in $K$ is also unramified in $K'.$
Finally, $KL=K'L$ and $K'$ has prime power degree over $\mathbb{Q}$.
I won't do the details. Essentially, all we did is obtain a new number field which also has prime power degree, where $q$ has no ramification and the primes unramified in $K$ are still unramified in $K'$. Therefore, repeating the procedure we obtain a number field of degree $p^k$ in which the only ramified prime is $p$.
Up to now I understood everything. But by a further reduction we can consider that $K$ is a cyclic extension, but I do not understand the proof, which is the following:
Since any abelian group is a direct sum of cyclic group of prime power order, any abelian extension is a composite of cyclic extensions of prime power degree. The proposition above allows us to assume that $p$ is the only ramified prime in each of these cyclic extensions. $\blacksquare$
I get that the abelian extension is the composite of cyclic extensions of prime power degree, but I do not understand why the above proposition guarantees that the number field $K'$ constructed above is still a cyclic extension of $\mathbb{Q}$. Any help?
EDIT:
User @reuns asked me to put more detail on the construction of $L$ and $K'$. Specifically, this is a little amount of exercises on Chapter $4$ of Marcus's Number Fields, and the specific exercise is the following:
