Make $3$-hypergraph Cayley on $ D_{2n} $ and $ \mathbb Z_n$ How we can make $3$-hypergraph Cayley on $ D_{2n} $ and  $ \mathbb Z_n$? 
Definition:
 let $G$ be a group, A $3$-hypergraph cayley on $G$ has a generator set $T$ with elements of order $3$ such that for $\sigma\in T$, $ \sigma^{-1}=\sigma^2\not\in T$ and $\{g_1,g_2,g_3\}\subseteq E(Cay_3(G,T))$ iff  $\exists ~\sigma\in T$ such that $g_2=g_1\sigma, g_3=g_2\sigma$. 
Thanks for advise.
 A: The first thing we need to know, is that to have elements of order $3$, we need the order of the group to be divisible by $3$.
For $\mathbb{Z}_n$, we need to have $n$ divisible by $3$.  So, assume $n=3m$.  The elements of order $3$ in $\mathbb{Z}_{3m}$ are those divisible by $m$, except the identity.  That is, the elements of order $3$ are $\pm m$.  So, according to the given definition, the only possible generating sets $T$ are $T=\emptyset$, $T=\{m\}$ and $T=\{-m\}$.
In the case $T=\emptyset$, any $3$-uniform hypergraph satisfies the property (since it's vacuous).  The $T=\{m\}$ and $T=\{-m\}$ cases will be identical, since $\langle m \rangle=\langle -m \rangle$ is used to generate the edges.
Here is an example for $\mathbb{Z}_{15}$:

For $D_{2n}$, we need to have $n$ divisible by $3$.  So, assume $n=3m$.  The elements of order $3$ in $D_{6m}$ are the "rotate by $m$" elements.  So if $$D_{2n}=\langle f,r | f^2=e, r^n=e, rf=fr^{-1} \rangle,$$ then the possible generating sets $T$ are $T=\emptyset$, $T=\{r^{2m}\}$ and $T=\{-r^{2m}\}$.  Like the cyclic case, the any $3$-uniform hypergraph fits into the $T=\emptyset$ case, and the $r^{2m}$ and $r^{-2m}$ cases are identical.
Here is an example for $D_{12}$, where $T=\{r^2\}$:

The different colors indicate different orbits under the group action.
