Lattice-complemented does not satisfy intermediate subgroup condition

Statement
It can happen that $$H$$ is a lattice-complemented subgroup of $$G$$, and $$H \le K \le G$$, with $$H$$ not a lattice-complemented subgroup in $$K$$.

Example of the symmetric group
Let $$G$$ be the symmetric group on a set $$\{ 1,2,3,4 \}$$ of size four.

Consider the subgroups:

$$H = \langle (1,3)(2,4) \rangle, \qquad K = \langle (1,2,3,4) \rangle$$.

$$H$$ is a subgroup of order two and $$K$$ is a subgroup of order four containing $$H$$. Further:


 * $$H$$ is a lattice-complemented subgroup in $$G$$: The symmetric group on the subset $$\{ 1,2,3 \}$$ is a lattice complement to $$H$$ in $$G$$.
 * $$H$$ is not a lattice-complemented subgroup in $$K$$: Indeed, $$K$$ is a cyclic group of order four and $$H$$ is a subgroup of order two, and has no lattice complements.