# Commutativity + “something” $\to$ associativity?

What additional properties must an operation have besides commutativity so that commutativity along with other properties implies associativity?

• Try "abstract algebra". en.wikipedia.org/wiki/Abstract_algebra – Frpzzd Oct 5 '17 at 23:35
• ...and it doesn't seem like commutativity paired with anything would imply associativity. Maybe it's just me, but commutativity and associativity seem pretty independent. – Frpzzd Oct 5 '17 at 23:36
• $(a,b)\mapsto \frac{a+b}{2}$ is commutative, but not associative at all. They are independent, otherwise it would make little sense to give the foundations of group and ring theory in the usual way. – Jack D'Aurizio Oct 6 '17 at 2:02

This is a reasonable question, but a reasonable answer (in my opinion) basically is negative: commutativity and associativity are not reasonably related, in most natural contexts.

The most disturbing-to-me example is that of Jordan algebras, which are commutative but not associative, despite being reasonably natural. A simple case is that of (real or complex) matrices of some fixed size, where the "Jordan product" is $a*b=ab+ba$.

For that matter, Lie algebras' anti-commutativity do suggest that associativity is a quite different thing ...

• Do you think that other way round is basically similar, that is, that associativity + "something" hardly can imply commutativity? What are your thoughts on this case? – user480281 Oct 5 '17 at 23:48
• Yes, I think that the two ideas, commutativity, and associativity, are somehow fundamentally different... – paul garrett Oct 6 '17 at 0:03
• Do you know of any expert when it comes to algebraic structures with whom I could talk about topics like this one? By e-mail correspondence. – user480281 Oct 6 '17 at 0:12

Here's why it's not possible, in most cases, to have that something. Commutativity deals with the inputs themselves. Associativity meanwhile, is all about parenthesis placement. The common part they have, is they are both defined for repeated operations, of the same type. Are we locked in by PEMDAS etc.? That's mostly a convention, to make sure people around the world, can look at the same arithmetic and get the same answer. So to really answer the question, as far as most can see here, they are completely unrelated. You can have associativity without commutativity, you can have commutativity without associativity I'm confident as well, you are asking for a property that is of their intersection, which may not be well defined.

• Hi Roddy. Welcome to MSE! – user480281 Oct 6 '17 at 1:00
• @AntoinedePaladin I've been here 4 months already, but thanks. – user451844 Oct 6 '17 at 1:04
• But you can be welcomed every day. – user480281 Oct 6 '17 at 1:07
• don't turn this site into facebook, also en.wikipedia.org/wiki/… may be a nice read for you as well as the permutations article. – user451844 Oct 6 '17 at 1:08

Suppose we have an operation $\star$ on a set $S$ such that for all $x,y,z\in S,$ we have $$x\star(y\star z)=(x\star z)\star y.$$ If $\star$ is also commutative, then $\star$ is associative.

The above is borrowed from Axiom 4 of Tarski's axiomatization of the real numbers. Axioms $4$ and $5$ together imply (and are implied by) the axioms of an abelian group: associativity, identity, inverses, and commutativity.