Commutative subtraction It is well known that subtraction is not commutative in general.
However, it is commutative in some groups: $\mathbb I$, $\mathbb C_2$, $\mathbb K_4$.
I am trying to understand the logic.
Considering a difference between elements $a$ and $b$ of a commutative magma (+) is an element $c$ of the magma with the following properties:


*

*$b + c = c + b = a$;  

*$c$ is unique;


subtraction can be defined as a binary operation on a commutative magma returning the difference between two elements.
Let's call a commutative magma subtractive if it is closed under subtraction.
Is there a simple way to find all subtractive magmas or groups with commutative subtraction?
Are there infinite subtractive magmas or groups with commutative subtraction?
 A: Suppose $S$ is a subtractive commutative magma in which subtraction is commutative.  Then for any $a,b\in S$, we have $$(a+b)+b=(a+b)+((a+b)-a)=(a+b)+(a-(a+b))=a$$ which means $a+b=a-b$.  Conversely, if $S$ is a subtractive commutative magma in which $a+b=a-b$ for all $a$ and $b$, then subtraction is obviously commutative.  So commutativity of subtraction is equivalent to subtraction being the same as addition.  Or, a commutative magma has subtraction which is commutative iff it satisfies the identity $(a+b)+b=a$ (since this identity implies subtraction exists and coincides with addition).
In particular, if $S$ is a group, subtraction being the same as addition just means that $a=-a$ or $2a=0$ for all $a\in S$.  Thus the abelian groups with commutative subtraction are exactly the abelian groups in which every element has order dividing $2$, i.e. the vector spaces over $\mathbb{Z}/2\mathbb{Z}$.
A bit more generally, suppose $S$ is a nonempty subtractive commutative semigroup.  Fix some element $a\in S$ and let $0=a-a$.  Then for any $b\in S$, $a+0+b=a+b$ and so $0+b=(a+b)-b=b$.  That is, $0$ is an identity element. We also see that $0-b$ is an inverse for $b$ and so $S$ is a group.  Thus the semigroup case reduces to the group case: a subtractive commutative semigroup with commutative subtraction is either empty or a vector space over $\mathbb{Z}/2\mathbb{Z}$.
For more general (not necessarily associative) magmas, I doubt there is any nice classification.  Here is one neat class of examples.  Let $(S,+)$ be any abelian group and fix an element $z\in S$.  Define an operation $\oplus$ on $S$ by $$a\oplus b=z-a-b.$$  I claim $(S,\oplus)$ is a commutative subtractive magma in which subtraction is commutative.  It is clear that $\oplus$ is commutative.  To see it is subtractive and subtraction is commutative, note that $a\oplus c=b$ iff $c=z-a-b$ and so the difference between $a$ and $b$ with respect to $\oplus$ is $z-a-b$.
A bit more generally, you could also take any subset of $S$ closed under this operation $\oplus$.  Note though that not every example arises in this way.  Indeed, any example of this form satisfies the identity $$((a\oplus b)\oplus c)\oplus d=((a\oplus d)\oplus c)\oplus b$$ (they are both equal to $z-a-b-d+c$).  However, this identity does not hold in every example; in particular, it does not hold in the free commutative magma satisfying the identity $(a+b)+b=a$ on four elements.  (Proof sketch: say that a word is reduced if it cannot be shrunken using $(a+b)+b=a$.  Prove by induction on length that each word can be reduced to a unique reduced word (modulo swapping the order of sums).  Conclude that each element of a free object is represented by a unique reduced word on the generators. Thus if $a,b,c,d$ are free generators, then $((a+b)+c)+d$ and $((a+d)+c)+b$ are distinct reduced words and so are not equal.)
