Non-Commutativity Implies Non-Associativity? I read this question, Does commutativity imply Associativity?.
And then was curious if non-commutativity implies non-associativity.
For concatenation (*), this is not true.
let A = {1,2}
let B = {3,4}
let C = {5,6}

A * B = {1,2,3,4}
does not equal
B * A = {3,4,1,2}
A * (B * C) = {1,2,3,4,5,6}

and
(A * B) * C = {1,2,3,4,5,6}
However, in general, does non-commutativity imply non-associativity?
Of course it's not true for concatenation, but perhaps for other algebras?
 A: The quaternions are a noncommutative (and associative) division algebra.
Commutativity and aasociativity are totally independent, even for structures as rigid as algebras, and it would be a good use of your time to stockpile some more examples.
The octonions are a noncommutative nonassociative division algebra.
The real numbers are a commutative and associative division algebra.
The cross product on triples of elements from a field of characteristic 2 form a nonassociative but commutative algebra.
Of course, moving to groups and magma makes everything even easier. Groups are all associative, but of course there are Abelian ones and nonabelian ones.
Someone mentioned my favorite nonassociative commutative magma already: the rock paper scissors magma, where rr=r, pp=p, ss=s, rs=r, rp=p, ps=s.
I don't have a favorite nonassociative noncommutative magma, but I imagjne it is very easy to haphazardly define multiplication on a small number of symbols to make one. Say, a, b, c with ab=c and ba=b, ca=c and the other products however you wish.
A: Taking A and B to be general NxN matrices we have a non-commutative algebra. One can construct a communitavie and non-associative algebra by defining op(A,B) to be (AB+BA)/2. A system that neither would be op(A,B)=(AB-BA)/2. In both cases, AB is ordinary matrix multiplication. The latter (AB-BA)/2 is sometimes called the Possonian is is used in abstract formulations of physics.
A: In particular, if this were true, then associativity would imply commutativity, by contraposition. This is clearly false, as your example shows.
