Permutably complemented does not satisfy transfer condition

This article gives the statement, and possibly proof, of a subgroup property (i.e., permutably complemented subgroup) not satisfying a subgroup metaproperty (i.e., transfer condition).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about permutably complemented subgroup|Get more facts about transfer condition|

Statement

It can happen that $H,K\leq G$ are subgroups such that $H$ is permutably complemented in $G$ but $H\cap K$ is not a permutably complemented subgroup in $K$.

Proof

Example of the dihedral group

Further information: dihedral group:D8

Let $G$ be 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:

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

Then, we have:

• $H$ is a permutably complemented subgroup of $G$: The subgroup $\{e,ax\}$ is a permutable complement to $H$ in $G$.
• $H\cap K$ is not a permutably complemented subgroup of $G$: $H\cap K$ is the group $\{e,a^{2}\}$, which clearly does not have any permutable complement inside $K$, which is a cyclic group of order four.