Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does commutativity imply associativity? I'm asking this because I was trying to think of structures that are commutative but non-associative but couldn't come up with any. Are there any such examples?

NOTE: I wasn't sure how to tag this so feel free to retag it.

share|improve this question
Commutative operations that are associative are the exception. But an important exception! Let $x\ast y=|x-y|$. –  André Nicolas Jun 20 '12 at 20:41
Not even in the presence of an identity element and an opposite, see this. In fact, William's answer is already in that post ;) –  lentic catachresis Jun 20 '12 at 21:58
The interchange law $(x * y) \cdot (z * w) = (x \cdot z) * (y \cdot w)$, in the presence of a two-sided common unit element, implies commutativity and associativity of $*$ and $\cdot$. (In fact, they have to be the same operation!) –  Zhen Lin Jun 20 '12 at 22:14
See my 3 February 2009 sci.math post A natural example of a commutative, non-associative operator (see Google archive version or Math Forum archive version) for some examples and references. –  Dave L. Renfro Jun 22 '12 at 21:31

4 Answers 4

up vote 27 down vote accepted

Consider the operation $(x,y) \mapsto xy+1$ on the integers.

share|improve this answer
Very nice example! –  Eugene Jun 20 '12 at 20:41

A basic example is the "midpoint" binary operation: $a*b = \frac{a+b}{2}$

In general, if $P(u,v)$ is any polynomial in two variables with rational coefficients, then $x*y = P(x+y,xy)$ is rarely associative - I'd be curious under what conditions on $P$ this operation would be associative.

My example is $P(u,v)=\frac{u}{2}$ and Marlu's example is $P(u,v)=1+v$.

share|improve this answer
In general, a symmetric function $P(x,y)$ is rarely associative. You could take $P(x,y)=\arctan xy+e^{(x+y)^9}$, too. –  Mariano Suárez-Alvarez Sep 14 '12 at 16:38

The easiest Jordan algebra is symmetric square matrices with the operation $$ A \ast B = (AB + BA)/2, $$ similar to a Lie algebra but with a plus sign.


share|improve this answer

Let $A = \{e,x,y\}$. Define $\cdot$ on $A$ to be $a\cdot e=a$ for all $a$, $e\cdot a= a$ for all a, and $a\cdot b=e$ for all $a$ and $b$ such that $a\neq e$ and $b\neq e$, (i.e. $a,b \in \{x,y\}$).

This operation is commutative, $e$ is the identity, (everything even has an inverse), but is not associative since $(x \cdot y) \cdot y = e \cdot y = y$ and $x \cdot (y \cdot y) = x \cdot e = x$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.