A subgroup H⊂GH \subset G is called core-free if for any normal subgroup N◃GN \triangleleft G then N⊂HN \subset H implies N=1N=1.

Of course, every proper subgroup of SS simple is core-free, and every core-free subgroup of AA 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.

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1 Answer

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Yes.

Assume that GG is a finite group wihtout non-trivial core-free subgroup. Let A1,…,ArA_1, \dots , A_r be the atoms of the subgroup lattice L(G)\mathcal{L}(G). Then by assumption, ∀i,Ai◃G\forall i, A_i \triangleleft G. If moreover L(G)\mathcal{L}(G) is atomistic then every subgroup of GG is normal, i.e. GG 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?

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Yes, for example ⟨x,y∣x3=y4=1,y−1xy=x−1⟩\langle x,y \mid x^3=y^4=1,y^{-1}xy=x^{-1} \rangle.

– Derek Holt

2 days ago

It should be the dicyclic group of order 12: sheaves.github.io/Subgroup-Lattice-Color-Vertices

– Sebastien Palcoux

2 days ago

Yes that’s right. Unfortunately I can never remember what dicyclic means!

– Derek Holt

yesterday

Wikipedia: << It is an extension of the cyclic group of order 22 by a cyclic group of order 2n2n, giving the name di-cyclic. >>

– Sebastien Palcoux

yesterday