# Permutably complemented is not finite-intersection-closed

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

It is possible to have a group $G$ and two permutably complemented subgroups $H, K \le G$ such that $H \cap K$ is not permutably complemented in $G$.

## Proof

### Example of the dihedral group

Further information: dihedral group:D8

Suppose $G$ is 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: $H = \langle a^2, x \rangle = \{ e, a^2, x, a^2x \}, \qquad K = \langle a^2, ax \rangle = \{ e, a^2, ax, a^3x \}$.

Then: $H \cap K = \{ e, a^2 \}$.

We have:

• $H$ is a permutably complemented subgroup of $G$: The subgroup $\{ e, ax \}$ is a permutable complement to $H$ in $G$.
• $K$ is a permutably complemented subgroup of $G$: The subgroup $\{ e, x\}$ is a permutable complement to $K$ in $G$.
• $H \cap K$ is not permutably complemented in $G$: This can be seen by direct inspection, but also follows from the more general fact that in a nilpotent group any nontrivial normal subgroup intersects the center nontrivially. Here, $H \cap K$ is the center, and if it has a permutable complement, that subgroup must be a nontrivial normal subgroup, leading to a contradiction.