# Are there "tri-commutative" structures for which: $AB \neq BA$, $BC \neq CB$, yet $ABC = BAC = ACB$?

Groups can be Abelian or non-Abelian, however I'm curious of the space between these two where there are either weaker forms of commutativity or special properties some elements have which endows the group with quasi-Abelian characteristics. Having a name for this could be handy for universal algebra.

Preferably a structure with associativity and where all products of three elements are commutative w.r.t. their neighbors (the elements to the right and left of them in the composition/sequence of the triple-product, so we don’t necessarily have $$ABC = CAB = CBA$$, but we can interchange $$A$$ with $$B$$ and $$B$$ with $$C$$ while preserving the outcome). There may be some non-Abelian groups which have specific subsets with this property, or even subgroups where all members have this property.

I'm under the impression there are theorems which hold for all groups which may be easy to prove for Abelian groups and hard to prove for non-Abelian groups. My intuition is that commutativity gives you a lot (at least when bundled with closure and associativity) and it might be tricky to prove something right at the edge of "things which are true for non-Abelian groups" if you could use commutativity as a shortcut. I'm not sure how well-founded this intuition, but clearly there are things which are common to all groups and things which are different between Abelian and non-Abelian groups; it's just whether one can use assumptions of commutativity instead of taking a longer path without it.

Binary relations are stronger than $$n$$-ary relations as the binary relations can imply the $$n$$-ary relations. For instance sets resulting from a Cayley-Dickson construction have power-associativity, which is much weaker than binary associativity because it only applies to $$x^n$$ rather than for products of distinct elements with $$n$$-many repetitions, or even that all products of $$n$$-many terms are associative but products of $$(n-1)$$-many terms are not necessarily so. To me power-associativity resembles a form of idempotency, however clearly in these algebras $$x^n \neq x$$. There may be some very interesting structures only possible with $$n$$-ary relations rather than binary relations, however my understanding is that if an algebra is pairwise commutative and tri-wise associative then it's also tri-wise commutative (we can permute $$ABC$$ however we like and preserve the result. Note: this doesn't imply commutativity between $$AC$$ if we have commutativity of $$\{A,B\}$$ and $$\{B,C\}$$, only that all triple-products of $$\{A,B,C\}$$ are the same).

Rings require commutativity of addition and fields require commutativity of both addition and multiplication, so clearly commutativity is important. There are non-commutative rings however addition is still commutative in these cases, so it would be interesting to see if there are many results for quasi-rings where addition is almost commutative, or how much commutative ring theory breaks down if we weaken the multiplicative commutativity. I'm aware of (Tropical) semi-rings however the condition which is weakened pertains to inverses and not to commutativity. While it isn't exactly weakening commutativity of fields there's Quantum Stochastic Calculus, it captures the spirit of the question which is to explore what happens when we weaken commutativity conditions.

A semi-group is too weak since:

1. We have closure
2. We have extra structure via our "not quite commutativity" properties

-Therefore I’m wondering if there are names for traits such as “Triad commutativity” or “Triad associativity”. Triad commutativity could be described as "A symmetric (w.r.t. it's arguments) trinary function which can be decomposed into (not necessarily symmetric) binary functions such that under composition these binary functions yield a symmetric trinary function". Triad associativity is analogous to the term power-associativity. There is $$n$$-ary associativity, so we could just have a set equipped with a binary operation and impose $$n$$-ary associativity on sufficiently long compositions of this binary operation. Power-associativity feels special compared to sets where associativity is arbitrarily restricted to a weaker form since it holds for Octonions, Sedenions, etc.

This resembles the concept of the center of a group, since we could look for a subgroup with "tri-commutativity" within a non-Abelian group. It might be messier to ask for a subset (not necessarily a subgroup, as we may lack closure) of a group where any two elements are "tri-commutative" with the rest of the group (or even more restrictive, to find the elements that are tri-commutative with any other two elements of the group, perhaps requiring these special elements to be the middle term in our triple product though this may not be necessary).

The question in the title is thus whether there are (named and hopefully interesting) structures for which $$g(x,y,z) = g(y,x,z) = g(x,z,y)$$ and $$f(x,y) \neq f(y,x)$$, $$f(y,z) \neq f(z,y)$$ [where $$g(a,b,c) = f(a,f(b,c)) = f(f(a,b),c)$$]. I'm not sure whether this implies $$g(x,y,z) = g(z,x,y) = g(z,y,x)$$, or if many non-commutative structures have this for the special case where for some $$x$$,$$z$$ we have $$f(x,z) = f(z,x)$$. Associativity can be represented as $$f(f(x, y), z) = f(x, f(y, z))$$.

Are there any names for things like these, such as “pseudo-Abelian", "sub-Abelian" or "hypo-Abelian”? Sub-Abelian might be a bad name since it could instead refer to Abelian subgroups of non-Abelian groups, such as how for any group $$G$$ and $$g\in G$$, then $$⟨g⟩=\{g^n:n\in Z\}$$ is Abelian. The subgroups generated this way seem very simple, but for some groups (e.g. Quaternions for $$\{±1\}$$) there might be much fancier ones to construct. Could we call the ability to carry out this construction "power sub-Abelian"?

• As a side note: For rings, the axiom that says "addition is commutative" is almost redundant. In fact, if we assume all ring axioms (except addition is commutative) and suppose that the ring has a multiplicative identity (i.e., a "1"), then addition is automatically commutative. Proof: 0x = (0+0)x = 0x+0x. Add -(0x) to both sides and get 0x=0. Next, 0=(-1+1)x=-1x+1x=-1x+x so -1x = -x (it is the additive inverse of x). Finally, -(-x+(-y))=-(-y)+-(-x)=y+x (i.e., the inverse of a+b is b inverse plus a inverse) and -(-x+(-y))=-1(-x+(-y))=-1(-x)+(-1)(-y)=-(-x)+(-(-y))=x+y. Therefore, x+y=y+x. Aug 10, 2020 at 13:32
• For -(-x+(-y))=-(-y)+-(-x)=y+x, how did you swap the order of x and y? In the first parentheses we have -x followed by (-y) [seperated by +] whereas after applying the distributive law you have (-y) followed by -1(-x) [seperated by +, and this time without parentheses]. Aug 10, 2020 at 14:30
• The reversal of order is sometimes referred to as the "socks-shoes principle". To undo something it must be undone in the reverse order. In multiplicative notation this looks like $(AB)^{-1}=B^{-1}A^{-1}$. In additive notation, we have "-(a+b)=-b+(-a)$. Aug 10, 2020 at 14:37 • Here's a version of the end of my argument with$a=-x$and$b=-y$:$-b+(-a)=-(a+b)$(socks-shoes)$=-1(a+b)$(using$-1x=-x$)$=-1a+(-1b)$(distributive law)$=-a+(-b)$(again using$-1x=-x\$). Aug 10, 2020 at 14:38

Such a structure wouldn't be able to have inverses, since if $$AB\neq BA$$ but $$ABC=BAC$$ then if $$C$$ has an inverse we get $$ABCC^{-1} = BACC^{-1} \Rightarrow AB=BA$$ which is a contradiction. So there are no groups with this property.