# 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?

• Subtraction is commutative in a group iff the group has exponent at most two. – Lord Shark the Unknown Mar 21 at 6:38

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.)