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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?

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    $\begingroup$ Subtraction is commutative in a group iff the group has exponent at most two. $\endgroup$ – Lord Shark the Unknown Mar 21 at 6:38
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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.)

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  • $\begingroup$ Awesome! Thank you. $\endgroup$ – Alex C Mar 21 at 11:28

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