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 $\langle H, K \rangle = G$.

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 $HK = G$.

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:

$H := \langle (1,3)(2,4) \rangle$.

• $H$ 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 $16$:

$G = \langle a,x \mid a^8 = x^2 = e, xax^{-1} = a^{-1} \rangle$.

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

$H = \langle a^4, x \rangle = \{ e, a^4, a^4x , x \}$.

• $H$ is a lattice-complemented subgroup of $G$: Indeed, the subgroup $\langle ax \rangle$ 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 $\langle a \rangle$ must have order either two or four. In either case, that intersection must contain the cyclic subgroup $\langle a^4 \rangle$. 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.