Is there a non-abelian group without non-trivial core-free subgroup?

Question

A subgroup $H \subset G$ is called core-free if for any normal subgroup $N \triangleleft G$ then $N \subset H$ implies $N=1$.
Of course, every proper subgroup of $S$ simple is core-free, and every core-free subgroup of $A$ abelian is trivial.

Question: Is there a non-abelian group without non-trivial core-free subgroup?

If necessary, we can assume all the groups to be finite.

Answer 1

Yes.

Assume that $G$ is a finite group wihtout non-trivial core-free subgroup. Let $A_1, \dots , A_r$ be the atoms of the subgroup lattice $\mathcal{L}(G)$. Then by assumption, $\forall i, A_i \triangleleft G$. If moreover $\mathcal{L}(G)$ is atomistic then every subgroup of $G$ is normal, i.e. $G$ is a Dedekind group. A non-abelian Dedekind group is called a Hamiltonian group. We see that any Hamiltonian group has the expected property. The first example is the quaternion group.

Bonus question: Is there a non-Dedekind group without non-trivial core-free subgroup?