A classification problem of the subsets in $\Bbb{Z} \times \Bbb{Z}$ which behave like odd numbers when considered in $\Bbb Z$ Classify all the subsets $S$ of the set of cartesian product of integers $\mathbb{Z} \times \mathbb{Z}$ such that: If $x$ is in $S$ then $-x$ also in $S$; if $x$ and $y$ are in $S$, then $x+ y$ is in complement of $S$ and finally if $x$ is in $S$ and $y$ is in the complement of $S$ then $x+y$ is in S.
 A: The subsets $S \subseteq \Bbb Z \times \Bbb Z$ that suffice


*

*$x \in S \Rightarrow -x \in S$  

*$x,y \in S \Rightarrow x+y \not \in S$ 

*$x \in S, y \not \in S \Rightarrow x+y \in S$.


are exactly the empty set and the complements of the subgroups of index $2$ in $\Bbb Z \times \Bbb Z$:
Let $S$ suffice all properties. We show that the complement $A:=S^c$ is an additive subgroup:


*

*suppose $0 \in S$, then $0 = 0+0 \not \in S$ by 2, a contradiction. So $0 \in A$.

*Let $x \in A$. Suppose $-x \in S$, then $x = -(-x) \in S$ by 1, a  contradiction. Thus $-x \in A$.

*Let $x,y \in A$. Then $x= (x+y)+(-y) \in A$ and as $-y \in A$ we conclude by 3 that $x+y \in A$.


Furthermore by 2 we know that in the quotient $\Bbb Z \times \Bbb Z /S^c$, which is well-defined as $S^c$ is a group, the sum of two non-trivial elements is $0$. As this implies that any two non-trivial elements are inverse to each other and inverses are uniquely determined, we have $\Bbb Z \times \Bbb Z /S^c = \Bbb Z /2 \Bbb Z$ or $\Bbb Z \times \Bbb Z /S^c = 0$, so $S^c$ is of index $1$ or $2$.
On the other hand, any complement of a subgroup of $\Bbb Z \times \Bbb Z$ suffices 1 and 3 and if its index is $1$ or $2$, the second condition holds as well.
A: If we define "$+$" on $\Bbb Z \times \Bbb Z$ as $(a,b)+(c,d)=(a+c,b+d)$ then $(\Bbb Z \times \Bbb Z,+)$ is a group.
Lemma 1: If S satisfy conditions $(1)$ and $(2)$ then $S^c$ is a subgroup of $\Bbb Z \times \Bbb Z$. (@benh already proved this.)
So you have to look in complements of subgroups of $\Bbb Z \times \Bbb Z$ that satisfy the $(2^{nd})$ condition.
Lemma 2: If $m\Bbb Z \times n\Bbb Z$ is a subgroup of $\Bbb Z \times \Bbb Z$ that its complement satisfies the second condition then its index in $\Bbb Z \times \Bbb Z$ is $1$ or $2$. (@benh already proved this.)
Lemma 3: For any $m,n\in\Bbb N$, the function $f:\Bbb Z_{mn}\rightarrow\frac{{\Bbb Z \times \Bbb Z}}{{m\Bbb Z \times n\Bbb Z}}$, defined by $f(x)=x+m\Bbb Z \times n\Bbb Z$. is an isomorphism, so
$$[\Bbb Z \times \Bbb Z:m\Bbb Z \times n\Bbb Z]=mn.$$
Theorem: The only subsets $S$ of $\Bbb Z \times \Bbb Z$ that satisfy those three conditions are,
$$(\Bbb Z \times \Bbb Z)^c=\emptyset,(2\Bbb Z \times \Bbb Z)^c,(\Bbb Z \times 2\Bbb Z)^c.$$

It seems that @Nate in his/her comment mentioned the answer very shortly!
A: We need three rules:


*

*$x\in S \implies -x\in S$

*$x\in S,y\in S \implies x+y\notin S$

*$x\in S, y\notin S \implies x+y\in S$


We can infer some attributes, like:


*

*$0 \notin S$, since both $x$ and $-x$ in $S$ and $0=x+(-x)$. Also, otherwise $0+y$ and $y$ must be in diffferent groups.

*$x\in S \implies x+x=2x\notin S$

*$x\in S \implies x+2x=3x \in S$

*In general, for any integer $n$, $2nx\in S$ and $(2n+1)x\notin S$.

*finally, if you have $\{x_1,x_2,...,x_n\}$ all in S, $\sum a_ix_i$ is in S iff $\sum a_i$ is odd.


So any subset S such that is defined by a "basis" $\{x_1,x_2,...,x_n\}$. The basis need to be lineary independent over $Z$, or it will be contradiction or not minimal. Infinite basis is also possible.
Proof that all 3 axioms holds:


*

*$x\in S \implies x=\sum a_i*x_i, \sum a_i \space odd \implies -x=\sum (-a_i)*x_i, \sum (-a_i) \space odd \implies-x\in S$

*$x\in S,y\in S \\
\implies x=\sum a_i*x_i, \sum a_i \space odd, y=\sum b_i*x_i, \sum b_i \space odd \\
\implies x+y=\sum(a_i+b_i)x_i, \sum(a_i+b_i)=(\sum a_i+\sum b_i) \space even\\
\implies x+y\notin S$

*similar proof for the third property.

