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$\begingroup$

As we know that given a vector space $E$, we can remove the scalar multiplication operator, then the elements of $E$ forms a group and commute with each others (we refer this commutative group to abelian group).

I am not 100% sure that if a vector space $E$ is a subset of the abelian group? Could anyone give me an answer and explanation please? Thanks in advance!

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    yes, first 4 conditions is abelian group – 2017-02-27
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    The word "subset" is misleading. If $V$ is a vector space, then $(V,+)$ is an abelian group. So it is $V$ itself (which is a subset of $V$, but ...). – 2017-02-27
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    Thanks@DietrichBurde. Would you mind explaining more detail about "So it is $V$ itself, which is a subset of $V$" please? Did you mean that given a vector space $V$, the abelian group $(V, +)$ of the vector space $V$ is a subset of $V$? – 2017-02-27
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    Yes, every set is a subset of itself. – 2017-02-28
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    Thank you @DavidWheeler. Actually, I am confused now. Regarding the vector space $V$, what's the relationship between $V$ and its abelian group $(V, +)$? I thought $(V,+)$ is a larger set than vector space $(V, \mathbb{F}, +, \cdot)$. – 2017-02-28
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    We say for any set $V$ that $V$ is a subset of $V$. But we would not say that $(V,+,\cdot)$ is a *subset* of $(V,+)$. Is that the question? – 2017-02-28
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    Thanks @GEdgar. I understand now. – 2017-02-28

1 Answers 1

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A vector space is composed of 4 things, a set $V$, a field $\Bbb F$, an operation $+$ and a scalar multiplication $\cdot$, written as $(V,\Bbb F,+,\cdot)$ here. It satisfies the following axioms

  1. $+$ is closed
  2. $+$ is commutative
  3. $+$ has inverses
  4. $+$ has an identity element
  5. $\cdot$ is a function from $\Bbb F\times V$ to $V$
  6. $\cdot$ is distributive over $+$
  7. $\cdot$ has an identity
  8. $\cdot$ is associative

The first 4 axioms means that $(V,+)$ is an abelian group, but notice that it's the same set, $V$ that we are using both for the abelian group and the vector space. The vector space adds a field which gives scalar multiplication that results in another element in $V$ and not in something else.