Lattice-complemented not implies permutably complemented

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., lattice-complemented subgroup) need not satisfy the second subgroup property (i.e., permutably complemented subgroup)
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Statement

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

Definitions used

Lattice-complemented subgroup

Further information: 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 $⟨ H, K ⟩ = G$ .

Permutably complemented subgroup

Further information: 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 $H K = G$ .

Proof

Example of the alternating group

Further information: alternating group:A4

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

$H : = ⟨ (1, 3) (2, 4) ⟩$ .

$H$ 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

Further information: dihedral group:D16

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

$G = ⟨ a, x ∣ a^{8} = x^{2} = e, x a x^{- 1} = a^{- 1} ⟩$ .

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

$H = ⟨ a^{4}, x ⟩ = {e, a^{4}, a^{4} x, x}$ .

$H$ is a lattice-complemented subgroup of $G$ : Indeed, the subgroup $⟨ a x ⟩$ 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 $⟨ a ⟩$ must have order either two or four. In either case, that intersection must contain the cyclic subgroup $⟨ a^{4} ⟩$ . 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.