Lattice-complemented is not transitive

Statement
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

 * Permutably complemented is not transitive

Example of the dihedral group
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.