Minimal Abstract Algebra

This is a challenge !!!!!

There are three objects, A, B, and C, and an operation called “doesn’t have a name”.
Two are similar, and the third is a bit different
They are paired to yield a single object as follows:

AA = BB = CC = C
AB = BC = CA = B
AC = CB = BA = A

Notice that BC and CB are different, so non-abelian.
Worse is that (AC)A = C and A(CA) = B are different, so non-associative.
And consequently A and B and C are different.

Your job is to identify (model, in current jargon) the objects and the operation.”

The next post will be “the solution”.

1 Comment

Filed under abstract, algebra, Uncategorized

One response to “Minimal Abstract Algebra

  1. I was thinking of some simple matrices and multiplication, like Pauli’s spin matrices, but then C would have to be the identity matrix and the operation would have to be associative.
    But I would still try with some matrices: These are three equations for three matrices, so for arbitrary 2×2 matrices there would be 12 equations in total for 12 numbers (or everything doubled for complex numbers). If something still does not work out right I would add one more dimension and try to solve 27 equations for 27 numbers. But even in these cases, associativity should hold – so matrix multplication is the wrong guess, and probably something odd happens when trying to solve these equations….

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