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.



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?

  • 3
    $\begingroup$ Yes, for example $\langle x,y \mid x^3=y^4=1,y^{-1}xy=x^{-1} \rangle$. $\endgroup$ – Derek Holt Oct 21 '16 at 20:41
  • $\begingroup$ It should be the dicyclic group of order 12: sheaves.github.io/Subgroup-Lattice-Color-Vertices $\endgroup$ – Sebastien Palcoux Oct 21 '16 at 23:58
  • $\begingroup$ Yes that's right. Unfortunately I can never remember what dicyclic means! $\endgroup$ – Derek Holt Oct 22 '16 at 7:53
  • $\begingroup$ Wikipedia: << It is an extension of the cyclic group of order $2$ by a cyclic group of order $2n$, giving the name di-cyclic. >> $\endgroup$ – Sebastien Palcoux Oct 22 '16 at 22:13

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