This question comes from this question by user72870. I shall explain how it relates to that question at the end.
Let me shortly define my question:
We call an embedding problem a diagram of the form:
$$\begin{matrix}&&\mathfrak G\\&&\downarrow\\G&\rightarrow&\Gamma\end{matrix},$$ where $G\rightarrow\Gamma$ is a group extension (therefore we assume that the homomorphism is surjective), and $\mathfrak G$ is the absolute Galois group of a field $K.$ See On Solvable Number Fields by J.Neurkirch for a reference.
And a solution of an embedding problem is a group homomorphism $\mathfrak G\rightarrow G$ making the above diagram commutative. Also see the linked paper for its field-theoretical meanings.
And my question is
If the kernel of $G\rightarrow\Gamma$ is cyclic, then is the embedding problem solvable for every $\mathfrak G$?
And if we require further that $\Gamma$ itself be cyclic?
The linked paper by Neukirch employs of group cohomology to prove the results, but, unfortunately, they always put some restraints on the order of the kernel, and on the field $K,$ which makes these results not applicable in my considerations, to be explained below. But they didn't consider the case where the kernel is cyclic, nor do they mention a reference about this case; I couldn't find anything useful on the internet, either.
Thus any help will be appreciated, thanks in advance.
Motivation:
The original problem is to prove that $(\mathbb Z/p^n\mathbb Z)^*$ is cyclic by embedding it in the multiplicative group of a field, which I failed to achieve without the cyclic assumption... Then I consider a diagram of the above form, where the group extension is $1\rightarrow (\mathbb Z/p^{n-1}\mathbb Z)^*\rightarrow(\mathbb Z/p^n\mathbb Z)^*\rightarrow(\mathbb Z/p\mathbb Z)\rightarrow1:$ the first map sends $x$ to $x\pmod p^n,$ and the second sends $y+p^{n-1}z, y\in(\mathbb Z/p^{n-1}\mathbb Z)^*, z\in(\mathbb Z/p\mathbb Z)$ to $z.$ Then I can find an extension $L\mid K$ with galois group $\cong \mathbb Z/p\mathbb Z.$ So, if this embedding problem is solvable, then there is an extension $E\mid K$ with Galois group $\cong (\mathbb Z/p^n\mathbb Z)^*;$ all that is left then is to take $K$ as a finite field, for finite extensions of a finite field are cyclic.
Even though this does not produce an embedding into the multiplicative group of a field, this still shows that our group in question is cyclic.
Thanks again for the attention.
