In M. Isaacs' Finite Group Theory, there is a remark on a problem which the author didn't provide any more details. The problem is this: Given a surjective group homomorphism $\theta:G\rightarrow G/N$, where $G$ is a finite group. If $P/N$ is a Sylow $p$-subgroup of $G/N$, then $P/N=\theta (Q)$ for some Sylow $p$-subgroup $Q$ of $G$. The proof is easy, & it depends on the fact that a Sylow $p$-subgroup does exist.
Now, the author remarked that the problem would not remain true if 'Sylow $p$-subgroup' were replaced by 'Hall $\pi$-subgroup', where $\pi$ is a prime set. However, I don't know where to find a counterexample. If one attempts to imitate the proof as in the Sylow case, one would know that the problem would remain true provided $G$ is solvable. (Since a solvable group has a Hall $\pi$-subgroup for every prime set $\pi$). So if $G$ is a counterexample, $G$ must be nonsolvable.
I'm looking at building $G$ from nonabelian simple groups, which I'm awfully unfamiliar with. So far I haven't got any useful results. Any idea what the minimal counterexample is?
Edit: @SteveD mentioned a counterexample in the comment. But I think I want a counterexample in which whenever we say the Hall $π$-subgroups of a group, we actually require the members of $π$ to divide $|G|$. Perhaps it's only slightly more difficult to figure out such a counterexample.
I think the remark only holds in that primes might vanish in the quotient, and the image thus is not any longer a $\pi$-subgroup:
Suppose that $H\le G$ is a $\pi$-Hall subgroup. Let $A=HN/N$ the image of $H$ in the factor. Then $A\le G/N$ is a subgroup such that $[G/N:A]$ is coprime to $\pi$, and $|A|=[HN:N]=[H:H\cap N]$ is a product of primes in $\pi$. It thus is a Hall subgroup, albeit might be so only for a subset of $\pi$, if certain primes do not divide the order of $G/N$.