# Lattice-complemented is not transitive

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

It can happen that we have subgroups $H \le K \le G$ such that $H$ is a lattice-complemented subgroup of $K$ and $K$ is a lattice-complemented subgroup of $G$ but $H$ is not lattice-complemented in $G$.

## Proof

### Example of the dihedral group

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

We have:

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