What assumptions are needed to prove that $(ab)c=b(ac)$ implies commutativity & associativity?

Question

The question is in the title: I would like to know what extra assumptions (if any) are needed for the following derivation that $(a\cdot b)\cdot c=b\cdot(a\cdot c)$ implies commutativity, i.e. $a\cdot b=b\cdot a$, and associativity, i.e. $(a\cdot b)\cdot c=a\cdot (b\cdot c)$.

If I assume the existence of an identity satisfying $a\cdot1=a$ for all $a$ in consideration, then $$a\cdot b=(a\cdot b)\cdot 1=b\cdot(a\cdot 1)=b\cdot a,$$ so we have commutativity, and given commutativity, we get $$(a\cdot b)\cdot c=(b\cdot a)\cdot c=a\cdot(b\cdot c),$$ which is associativity.

Are there any structures with no identity element such that $(a\cdot b)\cdot c=b\cdot(a\cdot c)$ for all $a,b,c$, but the operation is not commutative? Can other axioms be used in place of an identity (like, say, a cancellation law $a\cdot c=b\cdot c\implies a=b$)?

Answer 1

Consider matrices with even entries modulo $8$. Any product of three such matrices is zero, but we do not have commutativity.

Answer 2

To expand on Martin Bradenburg’s comment : define recursively a Carneiro expression on $\lbrace a,b \rbrace$ as follows :

1) $a$ and $b$ are themselves Carneiro expressions.

2) If $x$ and $y$ are Carneiro expressions, then $A(x,y)$ is a Carneiro expression.

3) Any Carneiro expression can be obtained by applying operations 1 and 2 a finite number of times.

Thus, $A(A(A(a,b),a),A(A(a,a),A(b,b)))$ is an example of a Carneiro expression. Denote by $C(a,b)$ the set of all Carneiro expressions on $\lbrace a,b \rbrace$. There is a unique map $d : C(a,b) \to {\mathbb N}$, such that $d(a)=d(b)=1$ and $d(A(x,y))=1+{\sf max}(d(x),d(y))$ for any $x,y\in C(a,b)$ : we call $d(w)$ the depth of $w$, for a Carneiro expression $w$.

Say that two Carneiro expressions $w_1$ and $w_2$ are elementarily equivalent if there are Carneiros expressions $x,y,z$ such that when we replace $A(A(x,y),z)$ by $A(y,A(x,z))$ somewhere in the expansion of $w_1$ (or $w_2$), then we obtain $w_2$ ($w_1$). Say that two Carneiro expressions $w_1$ and $w_2$ are equivalent if there is a finite sequence starting with $w_1$ and ending with $w_2$, such that each term in the sequence is elementarily equivalent to the next one. This is an equivalence relation, and we will denote it by $\sim$.

Say that a Carneiro expression is reduced if if contains no subexpression of the form $A(A(...,...),...)$. By induction on $d(w)$, any Carneiro expression is equivalent to a unique reduced Carneiro expression, that we denote by $r(w)$.

Let $R(a,b)$ denote the set of reduced Carneiro expressions on $\lbrace a,b \rbrace$. Define a binary operation $*$ on $R(a,b)$ by $x*y=r(A(x,y))$ for $x,y\in R(a,b)$.

Then, by construction, this binary operation satisfies your axiom $(ab)c=a(bc)$, also satisfies cancellation, but is not commutative (indeed $a*b \neq b*a$) nor associative (indeed $(a*b)*a$ and $a*(b*a)$ are not equal, since the first one computes to $A(b,A(a,a))$ and the other computes to $A(a,A(b,a))$. )