Let $G$ be a group, $G'\le G$ a subgroup, and $\varphi:G'\to H$ a surjective group homomorphism. Must there exist a surjective group homomorphism $\psi:G\to H'$ such that $H'\ge H$?
I know the converse is true: Take $G'=\psi^{-1}(H)$ and let $\varphi=\psi$.
I know if $G$ is abelian then the question is true: Take $H'=G/\ker(\varphi)$.
Whether the full direction asked above is true or false is unclear. I have asked some friends for their help and we couldn't reach a conclusion.
If anyone can point to a reference with a full solution to the problem, that is also great. Thanks in advance!
Your question is basically: does every subquotient of $G$ embed into a quotient of $G$?
In other words, if $K \trianglelefteq H \leq G$, does there exist $N \trianglelefteq G$ such that $H/K$ is isomorphic to a subgroup $G/N$?
The answer is no in general, but I cannot think of any examples which would be easy to describe. A computer calculation shows that smallest finite examples occur for $|G| = 64$, for example SmallGroup(64,18) has SmallGroup(16,13) as a subquotient, but not as a subgroup of a quotient.
In the case where $G$ is simple the question is whether every subquotient of $G$ is isomorphic to a subgroup of $G$. Here too you can find many examples. For example the Mathieu group $G = M_{24}$ has $A_6$ has a subquotient, but not as a subgroup.
There are probably easier examples, which perhaps others could suggest.