I have gotten the next structures: $$M_1=\langle \mathbb{N},S^{M_1}=\{\langle n,k\rangle\in \mathbb{N}^2:\exists m\in \mathbb{Z}(n-k=2m) \} \rangle$$ $$M_2=\langle \mathbb{Z},S^{M_2}=\{\langle a,b\rangle\in \mathbb{Z}^2:\exists m\in \mathbb{Z}(a-b=2m) \} \rangle$$ $$M_3=\langle \mathbb{Q},S^{M_3}=\{\langle x,y\rangle\in \mathbb{Q}^2:\exists m\in \mathbb{Z}(x-y=2m) \} \rangle$$
Now, I have the find the isomorphic structures.
In my opinion, there isn't isomorphism at all between the next structures, because there are negative numbers at $M_2$ and in $M_3$ there are fractions numbers whose difference is an integer and an even number, so I don't see a bijective function between them. I will be glad for some help. Thanks!
$M_1$ and $M_2$ are equivalence relations, where all even integers become equivalent, and likewise for odd integers. So can you find a bijection between $\Bbb N$ and $\Bbb Z$ that preserves parity?
In $M_3$ we also have an equivalence relation, but many more classes.... Can you show that isomorphisms preserve the number of classes under a relation? Or note that in $M_1,M_2$ the statement
$$\forall x,y,z \in M: xSy \lor xSz \lor ySz$$
holds in $\langle M,S\rangle$ and in $M_3$ it does not..