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Is it possible to construct a quasi-vectorial space without an identity element?

I know that the commutativity of vector addition is a consequence of the others conditions in the definition of a vector space. I do not think that the associativity of vector addition is a consequence of the others conditions (I am including the commutativity among them). How do I prove that? In fact, I do not even know what I should do.

I would appreciate any help.

  • $\begingroup$ A ${\it vector\ space}$ is an additive group with some additional structure. An ${\it additive\ group}$ is a commutative group where the group operation is denoted by a $+$-sign. In any group the group operation is associative by definition. In short, in a vector space the addition is associative by definition. $\endgroup$ Jan 29, 2012 at 20:41
  • $\begingroup$ See this previous question for a discussion of the independence of the vector space axioms. $\endgroup$ Jan 29, 2012 at 21:51
  • $\begingroup$ @Christian, I beleive the point of the question is, if something satisfies all the vector space axioms except possibly associativity of addition, can it be proved that in fact it does satisfy associativity of addition? $\endgroup$ Jan 29, 2012 at 23:14
  • $\begingroup$ Commutativity only follows form the other axioms if the additive inverse axiom (equivalently, the cancellation axiom) is stated as holding on both sides. In most axiomatizations of vector spaces that I am familiar with in which the axioms of vector addition are stated explicitly (as opposed to simply saying "$(V,+)$ is an abelian group"), this is not the case: both the additive identity and additive inverse axioms are stated on a single side; if you do that, then commutative does not follow from the other axioms, as shown in the examples given in the post refered to above. $\endgroup$ Jan 29, 2012 at 23:20


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