Let $\phi\colon G\rightarrow H$ be a surjective homomorphism between finite groups. Assume that $\phi$ is not split, in other words there exists no homomorphism $\sigma\colon H\rightarrow G$ such that $\phi \circ \sigma = \text{Id}$.
Question: does there exists an $H$-extension of $\mathbb{Q}$ that is not contained in a $G$-extension?
Some remarks:
- If $A$ is a finite group and $k$ a field, an $A$-extension of $k$ is by definition a Galois extension $K/k$ with Galois group $A$.
- You may assume various strong forms of the inverse Galois problem over $\mathbb{Q}$, because I'm just trying to figure out whether this statement is true or not.
- The answer is affirmative when $\phi$ equals the quotient map $\mathbb{Z}/p^2\mathbb{Z} \rightarrow \mathbb{Z}/p\mathbb{Z}$ if $p$ is prime: take a different prime $\ell$ such that the size of $\mathbb{F}_{\ell}^{\times}$ is divisible by $p$ but not $p^2$. Then one can check using the Kronecker-Weber theorem that the unique $\mathbb{Z}/p\mathbb{Z}$-extension contained in $\mathbb{Q}(\zeta)$ (where $\zeta$ is an $\ell$-th root of unity) is not contained in a $\mathbb{Z}/p^2\mathbb{Z}$-extension. I think a similar argument works when $G$ (and hence $H$) is abelian.
- The answer to this question can be no when $\mathbb{Q}$ is replaced by another number field. Indeed, if $k = \mathbb{Q}(i)$ then every quadratic extension lifts to a $\mathbb{Z}/4\mathbb{Z}$-extension by Kummer theory.
- On the other hand, I would also be interested in the weaker question of whether there exists some number field $k$ which admits an $H$-extension that is not contained in a $G$-extension. There might be more hope to solve this unconditionally.
Any help and comments are appreciated!
