There may be multiple subgroups that are pairwise permutable complements

Statement

Let $n$ be any natural number. Then, there exists a finite group $G$ with subgroups $H_{1},H_{2},\dots ,H_{n}$ such that $H_{i}$ and $H_{j}$ are Permutable complements (?) in $G$ for any $i\neq j$.

Proof

Let $p$ be a prime such that $p+1\geq n$. Let $G$ be the group $\mathbb {Z} /p\mathbb {Z} \times \mathbb {Z} /p\mathbb {Z}$: in other words, $G$ is an elementary Abelian group of order $p^{2}$. $G$ has $(p+1)$ cyclic subgroups of order $p$, and any two of these are permutable complements.