Suppose we have $E=\{\mathrm{Car}\}$ with the following operations: $\mathrm{Car}+\mathrm{Car}=\mathrm{Car}$ and $\lambda\cdot \mathrm{Car}=\mathrm{Car}$ for all $\lambda \in \mathbb{R}$. I'm wondering if this could be a commutative group or a vector space.
Attempt: seems to be a structure with close "addition", i.e., $\forall a,b\in E, \ a+b\in E$, also it's associative, commutative and its unique element $\mathrm{Car}$ should be the neutral element.
So I think it should be an abelian group, but then the condition of $\lambda\cdot \mathrm{Car}=\mathrm{Car}$ makes me think there is more than one neutral element (in $\mathbb{R}$) which is not possible, same for Vector space, since the neutral element in $\mathbb{R}$ will not be unique. And end it up accepting this couldn't be vectorial space nor abelian group. It's ok?
Best Answer
Short version: In the formulation of the axioms for a group or vector space, the (multiplicative) identity element need not be unique, so this is a nonissue and $E$ may be claimed as a group with $(E,+)$ and real vector space with $(E,+,\cdot)$, a trivial one in both respects.
Long version:
So the crux of the issue seems to be the bolded axiom below, in the formulation of vector spaces:
This element need not be unique, so it is no issue.
Yes, $\mathbb{R}$ has its own, unique multiplicative identity. (Indeed, any group has an identity element, and one can prove that this element is unique, although our axioms - like above - only the posit the existence of at least one.) However, that is in reference to the structure $\mathbb{R}$ by itself; it need not hold for any vector space dependent on $\mathbb{R}$ (and indeed, for this context, we don't even care about the uniqueness of the scalar-multiplication identity).
Hence why you have comments saying $E$ is (up to isomorphism) the trivial vector space, the one consisting of only a zero vector.
In fact, we can say more: for any nontrivial $\mathbb{F}$-vector space $V$, then the $1\in \mathbb{F}$ in terms of the field axioms and the $1 \in \mathbb{F}$ in terms of the vector space axioms are always one and the same. Let us call these $\f,\v$ respectively.
Of course, you have seen how this breaks down for the trivial vector space. (I suppose one could go on about why we consider this a vector space, then, but it's largely due to convenience of including it in many results - somewhat analogous to why we might exclude the trivial ring as a field in certain definitions of the field axioms, since it's an exception to a number of results in field theory, despite otherwise meeting the axioms.)