# Thread: Combining the associative and commutative property into one

1. Can anybody do better than this?
Say I define a specific type of algebraic structure (I guess, a commutative monoid) with the following axioms:
1) There exists an identity element in the structure (G) e such that for all a in G, ae = a.
2) For all a, b, and c in G, (ab)c = b(ac)
And from only these we can deduce the commutative and associative axioms!
ab
= (ab)
= (ab)e (identity axiom)
= b(ae) (axiom #2)
= b(a) (identity axiom)
= ba
And thus ab = ba. And now for the associative property,
(ab)c
= (ba)c (commutative property)
= a(bc) (axiom #2)
Is there any statement that would deduce both the commutative and associative properties without the need of an identity element?

2.

3. What about in the case of a non-abelian group? Then you would need the associative axiom without the commutative axiom.

4. What about in the case of a non-abelian group? Then you would need the associative axiom without the commutative axiom.
The special axiom that (ab)c = b(ac) is not meant to replace the associative axiom or the commutative axiom, but both if they are in fact both true. The special thing about both being true is that you can arrange the terms and parenthesis any way you like and this makes things so much easier.
They always say that when constructing a system (such as the system of real numbers with addition and multiplication), assume as little as possible.

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