# Is this a group or a vector space?

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?

• A vector space is an additive abelian/commutative group with extra structure May 15 at 20:44
• A group is a set with respect to an operation, in this case $+$. May 15 at 20:48

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:

Definition: A set $$V$$ - over the field $$\mathbb{F}$$, with operators of $$+ : V^2 \to V$$ and $$\cdot : \mathbb{F} \times V \to V$$ - is a vector space if and only if it satisfies the axioms below:

• ... [omitted for brevity] ...
• There is an element $$1 \in \mathbb{F}$$ such that $$1x=x$$ for all $$x \in V$$

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.

Claim: Let $$V$$ be an $$\mathbb{F}$$-vector space where $$|V| \ge 2$$, and $$\f \cdot x = \v \cdot x$$ for each $$x \in V$$. Then $$\f=\v$$.

Proof: We work by contradiction, i.e. suppose that $$\f,\v$$ have the property of $$1x=x$$ for each $$x \in V$$ and $$1 \in \{\f,\v\}$$, but $$\f \ne \v$$.

Take $$x \in V \setminus \{0\}$$. Set $$\f \cdot x = \v \cdot x$$. Consequently, $$(\f-\v) \cdot x = 0$$. Then, multiplying the by the inverse of our coefficient, $$x = \frac{1}{\f-\v} \cdot 0 = 0$$ a contradiction.

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.)

It is the trivial vector space. If you ignore scalar multiplication, then it is the trivial group also.

• But can I ignore the scalar multiplication? i.e., doesn't change something? May 15 at 20:53
• There's nothing stopping you, @Valent, except that, to be precise, to ignore scalar multiplication, you use what is known as a forgetful funtor. May 15 at 20:55