Not containing $ 2^{i}$-th primitive roots of unity in cyclic galois extension of number field

Question

Suppose $K$ is an algebraic number field. We want to determine whether, for all even $n$, there always exists a cyclic Galois extension $L$ of degree $n$ over $K$ such that the intermediate field $K_2$ does not contain any primitive $2^i$-th root of unity for $i \geq 3$ where $ K _2 $ is the intermediate field.

If $ K = \mathbb{Q} $ then it is true. Since for all $ n $ there exist infinitely many primes $ p $ such that $ n $ divides $ n -1 $. As $ p$-th cyclotomic polynomial is irreducible over $ \mathbb{Q} $ then using Galois corresponding theorem we get a cyclic galois extension $ L $ of degree $ n $. Now since $ \phi(2^{i}) > 2 $ where $ i \geq 3 $ so any primitive $ 2^i $-th root of unity can not lie in $ K_2 $.

Answer 1

It's not quite clear from the post what $K_2$ is, but I'll just pretend $K_2=L$, which gives the strongest result.

Those infinitely many extensions constructed over $\mathbb{Q}$ are all linearly disjoint (e.g. since in each one, the respective prime $p$ is the only ramified prime), and thus there are in particular infinitely many $L/\mathbb{Q}$ among them which are linearly disjoint to $K(\zeta_8)$. Then firstly $LK/K$ is still cyclic of degree $n$, and also $LK$ has no $2^i$-th roots of unity which weren't already in $K$. So yes, it's possible unless in the trivially impossible case where $K$ itself contains such roots of unity.