Lattice-complemented not implies permutably complemented

Statement
A lattice-complemented subgroup of a group need not be permutably complemented.

Lattice-complemented subgroup
A subgroup $$H$$ of a group $$G$$ is termed lattice-complemented in $$G$$ if there exists a subgroup $$K$$ of $$G$$ such that $$H \cap K$$ is trivial and $$\langle H, K \rangle = G$$.

Permutably complemented subgroup
A subgroup $$H$$ of a group $$G$$ is termed permutably complemented in $$G$$ if there exists a subgroup $$K$$ of $$G$$ such that $$H \cap K$$ is trivial and $$HK = G$$.

Example of the alternating group
Let $$G$$ be the alternating group on the set $$\{ 1,2,3,4\}$$. Consider the two-element subgroup:

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


 * $$H$$ is lattice-complemented in $$G$$: The cyclic subgroup generated by $$(1,2,3)$$ is a lattice complement to $$H$$ in $$G$$.
 * $$H$$ is not permutably complemented in $$G$$: This can be seen by inspection. In fact, there is no subgroup of order six in $$G$$, and any permutable complement to $$H$$ must have order six.

Example of the dihedral group
Let $$G$$ be the dihedral group of order $$16$$:

$$G = \langle a,x \mid a^8 = x^2 = e, xax^{-1} = a^{-1} \rangle$$.

Let $$H$$ be the subgroup of $$G$$ generated by $$a^4$$ and $$x$$:

$$H = \langle a^4, x \rangle = \{ e, a^4, a^4x, x \}$$.


 * $$H$$ is a lattice-complemented subgroup of $$G$$: Indeed, the subgroup $$\langle ax \rangle$$ is a lattice complement to $$H$$ in $$G$$.
 * $$H$$ is not a permutably complemented subgroup of $$G$$: Any permutable complement $$K$$ to $$H$$ must have order four. Hence, the intersection of $$K$$ with $$\langle a \rangle$$ must have order either two or four. In either case, that intersection must contain the cyclic subgroup $$\langle a^4 \rangle$$. Hence, $$K$$ does not intersect $$H$$ trivially.

A generic example: simple non-Abelian group
Every simple non-Abelian group is a K-group: every subgroup is lattice-complemented. On the other hand, by Hall's theorem on solvability, there does not exist a $$p$$-Sylow complement for every $$p$$, so not every $$p$$-Sylow subgroup is permutably complemented. Thus, we can always find a lattice-complemented subgroup that is not permutably complemented.