Which groups $G$ has the property that for all subgroups $H$ , there is a surjective map from $G$ to $H$?

Question

I tried many examples , but i can't find any counterexample .
But I guess there are many counter examples , and specific sorts of groups or subgroups have this property (e.g abelian groups or normal subgroups).
Thus I have two question:

1. Is there any counter example of group $G$ and its subgroup $H$ s.t there is no surjective homomorphism from $G$ to $H$ ?

   2. If exists some counterexamples , which sort of groups or subgroups have this property?

I would also appreciate any reference .

Answer 1

Just a convincing example to show that this property is very far from being natural : take the good old symmetric group $S_n$ (let's take $n\geqslant 5$ to be safe). Then any group of order $n$ is a subgroup of $S_n$, but there is no surjective morphism from $S_n$ to any of them. This is because the only surjective morphisms from $S_n$ to any group are the identity, the signature and the trivial morphism.

Answer 2

Take $G = F(\{x_1,\dots, x_n\})$, the free group on $n$ generators. The commutator subgroup $G' = [G,G]$ is a free group of infinite rank and thus $G$ cannot surject onto $G'$, as it simply doesn't have enough generators (it has finite rank).