Permutably complemented is not finite-intersection-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., permutably complemented subgroup) not satisfying a subgroup metaproperty (i.e., finite-intersection-closed subgroup property).
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Statement

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

Proof

Example of the dihedral group

Further information: dihedral group:D8

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:

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

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.