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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. $\endgroup$ – Brian Fitzpatrick Jul 27 '14 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 $x$,$x^{-2}=x^{-1}\cdot x^{-1}$ $\endgroup$ – 89085731 Jul 27 '14 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 $\mathbb{Z}/(6)$. $\endgroup$ – Alex Youcis Jul 27 '14 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. $\endgroup$ – Hubble Apr 20 '16 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. $\endgroup$ – M. Vinay Apr 21 '16 at 5:50
  • $\begingroup$ Great answer Hubble $\endgroup$ – biki Dec 14 '16 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). $\endgroup$ – cws Jul 27 '14 at 6:31
  • $\begingroup$ @cws Ah, that explains it. Thanks. $\endgroup$ – user_of_math Jul 27 '14 at 11:31

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