Consider an extension of groups $$(E)\ :\ 1\to N\to G\stackrel{\pi}{\longrightarrow} Q\to 1$$ and assume

1. $N$ is abelian,
2. $Q$ is finite,
3. for any Sylow subgroup $S_p$ of $Q$, the pullback $(E_p)\ :\ 1\to N\to \pi^{-1}(S_p)\to S_p\to 1$ is split.

Then $(E)$ is split.

[A proof can be written using only elementary homological algebra plus Cartan-Eilenberg double coset formula.]

If we drop the first hypothesis, the conclusion doesn't hold in general. The simplest example seems to be the extension $$1\to F_2\to PSL_2(\mathbf Z)\simeq \mathbf Z/2\ast\mathbf Z/3\to PSL_2(\mathbf F_2)\simeq\mathfrak S_3\to 1$$

Even if we insist that $N$ be finite solvable, there are counter-examples (e.g. take the above with $PSL_2(\mathbf Z/8)$ instead of $PSL_2(\mathbf Z)$).

Assume $N$ is non-abelian and $\pi$ is split over all normalizers of Sylow subgroups in $Q$. Is $\pi$ necessarily split ?

Consider an extension of groups $$(E)\ :\ 1\to N\to G\stackrel{\pi}{\longrightarrow} Q\to 1$$ and assume

1. $N$ is abelian,
2. $Q$ is finite,
3. for any Sylow subgroup $S_p$ of $Q$, the pullback $(E_p)\ :\ 1\to N\to \pi^{-1}(S_p)\to S_p\to 1$ is split.

Then $(E)$ is split.

[A proof can be written using only elementary homological algebra plus Cartan-Eilenberg double coset formula.]

If we drop the first hypothesis, the conclusion doesn't hold in general. The simplest example seems to be the extension $$1\to F_2\to PSL_2(\mathbf Z)\simeq \mathbf Z/2\ast\mathbf Z/3\to PSL_2(\mathbf F_2)\simeq\mathfrak S_3\to 1$$

Even if we insist that $N$ be finite solvable, there are counter-examples (e.g. take the above with $PSL_2(\mathbf Z/8)$ instead of $PSL_2(\mathbf Z)$).

Assume $N$ is solvable and $\pi$ is split over all normalizers of Sylow subgroups in $Q$. Is $\pi$ necessarily split ?