It looks to me like you are trying to prove something that is NOT TRUE. For example, the set of positive integers is closed under ordinary mulitplication which is associative and both cancellation laws hold. But this is not a group.
#5
adriank
534
1
Ah, indeed. And if you take the integers ≥ 2 under multiplication, then you don't even get a monoid. Apparently such a thing is called a cancellative semigroup.
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).
Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?