Permutably complemented is not transitive

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

Statement with symbols

It can happen that we have groups $H\leq K\leq G$ such that $H$ is a permutably complemented subgroup of $K$ and $K$ is a permutably complemented subgroup of $G$, but $H$ is not a permutably complemented subgroup of $G$.

Proof

Example of the dihedral group

Further information: dihedral group:D8

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

We have:

• $H$ is permutably complemented in $K$: The subgroup $\{x,e\}$ is a permutable complement to $H$ in $K$.
• $K$ is permutably complemented in $G$: The subgroup $\{ax,e\}$ is a permutable complement to $K$ in $G$.
• $H$ is not permutably 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 permutable complement to the center must be a nontrivial normal subgroup, and hence such a thing cannot exist in a nilpotent group.