Lattice-complemented is not transitive

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This article gives the statement, and possibly proof, of a subgroup property (i.e., lattice-complemented subgroup) not satisfying a subgroup metaproperty (i.e., transitive subgroup property).
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It can happen that we have subgroups H \le K \le G such that H is a lattice-complemented subgroup of K and K is a lattice-complemented subgroup of G but H is not lattice-complemented in G.

Related facts


Example of the dihedral group

Further information: dihedral group:D8, Klein four-subgroups of dihedral group:D8, center of dihedral group:D8, subgroup structure of dihedral group:D8

Let G be the dihedral group of order eight; specifically:

G = \langle a,x \mid a^4 = x^2 = e, xax^{-1} = a^{-1} \rangle.

Let H be the center of G: H = \{ a^2 , e \}.

Let K be the elementary Abelian subgroup generated by a^2 and x, so K = \{ e, a^2, a^2x, x \}.

We have:

  • H is lattice-complemented in K: The subgroup \{ x, e\} is a permutable complement to H in K, and in particular, a lattice complement to H in K.
  • K is lattice-complemented in G: The subgroup \{ ax , e\} is a permutable complement to K in G, and in particular, is a lattice complement to K in G.
  • H is not lattice-complemented in G: This can be seen by inspection, but it also follows from a more general fact about nilpotent groups: every nontrivial normal subgroup of a nilpotent group intersects the center nontrivially. A lattice-complement to the center must be a permutable complement, hence must be a nontrivial normal subgroup, and hence such a thing cannot exist in a nilpotent group.