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When comparing the difference between the definition of vector space, I see that the main job is that vector space defines a scalar product while the group not, so here list two of my questions?

1.Why we need to define a scalar product for a vector space? Physical sense or some insight behind it?

2.One truly nice thing for vector space is that we represent the element with basis, so what we do with elements in vector space is just with basis,so why we can't do the same thing for group?

I think the question may be a little silly, but I need a question.

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  • $\begingroup$ Not directly related to your listed questions, but you might find interesting the definition of an $R$-module where $R$ is a ring. A vector-space over a field $k$ is the same thing as a $k$-module while an abelian group is the same thing as a $\Bbb Z$-module. $\endgroup$ Commented Jul 27, 2014 at 4:37
  • $\begingroup$ @BrianFitzpatrick That's what I need. I just wondering if I can define a basis in the $\mathbb{Z}$ module. For example let $x^2=x\cdot x$,$x^{-1}$ as the inverse element of $x$，$x^{-2}=x^{-1}\cdot x^{-1}$ $\endgroup$ Commented Jul 27, 2014 at 4:57
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    $\begingroup$ It might be worth mentioning, if it is non-obvious, that there are (many!!) groups which cannot be given the structure of a vector space. For example, $\mathbb{Z}$, or $\mathbb{Z}/(6)$. $\endgroup$ Commented Jul 27, 2014 at 4:58

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As you may know, a vector space is a set $V$ together with operations $+:V \times V \to V$ and $\cdot:K \times V \to V$ that satisfy certain conditions, where $K$ is a field (take $K = \mathbb{R}$ for instance). Turns out that these conditions makes $(V,+)$ into an abelian group, a fancy term for a commutative group. This means that if you take $V$ and remove the scalar multiplication operator, the elements of $V$ forms a group and commute with each others.

Conversely, you can take an abelian group and try to turn it into a vector space by adding scalar multiplication on it. This additional structure comes in handy when you want to reason about lengths and angles of vectors in $V$. A geometric interpretation of this is that it stretches, or contracts, vectors $v \in V$ by a constant factor $\alpha \in K$. In fact, scalars scale vectors.

Without scalar multiplication, it is not possible think of any way of constructing a basis in a group $G$. If you think back of the definition of a basis, you will see that it involves a field. The definition of a vector space encapsulates the notion of basis in some sense.

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    $\begingroup$ @M.Vinay I meant for certain abelian groups only, not all of them, hence the use of the word try. Good catch though. $\endgroup$ Commented Apr 20, 2016 at 22:52
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    $\begingroup$ So, to summarize (for the reader of the answer): You can try to convert a (commutative) group into a vector space, but it might not be possible in some cases. Also, there may be several (non-equivalent) ways to build up vector spaces from the same group. For example, the additive group of real numbers, $(\mathbb R, +)$ can be a one-dimensional vector space over the field $\mathbb R$, or an (uncountably) infinite-dimensional vector space over $\mathbb Q$ (the rationals). The two (obviously non-isomorphic) spaces don't just share the same underlying set, but the same underlying Abelian group. $\endgroup$ Commented Apr 21, 2016 at 5:50
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    $\begingroup$ Great answer Hubble $\endgroup$ Commented Dec 14, 2016 at 23:48
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Defining scalar products in vector spaces allows you to introduce notions of angle between two vectors, and also gives a definition of the length of a vector.

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    $\begingroup$ A clash of language here: the OP is using "scalar product" to mean scalar multiplication, and you seem to be interpreting it as a synonym for "inner product" (which it often is). $\endgroup$ Commented Jul 27, 2014 at 6:31
  • $\begingroup$ @cws Ah, that explains it. Thanks. $\endgroup$ Commented Jul 27, 2014 at 11:31

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