Permutably complemented is not finite-intersection-closed

Statement
It is possible to have a group $$G$$ and two permutably complemented subgroups $$H, K \le G$$ such that $$H \cap K$$ is not permutably complemented in $$G$$.

Example of the dihedral group
Suppose $$G$$ is the dihedral group of order eight:

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

Consider the two subgroups:

$$H = \langle a^2, x \rangle = \{ e, a^2, x, a^2x \}, \qquad K = \langle a^2, ax \rangle = \{ e, a^2, ax, a^3x \}$$.

Then:

$$H \cap K = \{ e, a^2 \}$$.

We have:


 * $$H$$ is a permutably complemented subgroup of $$G$$: The subgroup $$\{ e, ax \}$$ is a permutable complement to $$H$$ in $$G$$.
 * $$K$$ is a permutably complemented subgroup of $$G$$: The subgroup $$\{ e, x\}$$ is a permutable complement to $$K$$ in $$G$$.
 * $$H \cap K$$ is not permutably complemented in $$G$$: This can be seen by direct inspection, but also follows from the more general fact that in a nilpotent group any nontrivial normal subgroup intersects the center nontrivially. Here, $$H \cap K$$ is the center, and if it has a permutable complement, that subgroup must be a nontrivial normal subgroup, leading to a contradiction.