Is linear space totally different from group? Linear space builds on Abelian group. My question is, is linear space TOTALLY different from group?
Is it true that some properties of linear space are the properties of the Abelian group?
Actually, I still can not give an example, but I find they are really similar in a sence, though there are also many difference between them.
 A: A linear space $V$ is commonly referred to as a vector space. Vector spaces are like groups in that two vectors have a commutative, associative addition with identity (and thus are an abelian group).
But they're special in that there is a sort of 'multiplication,' but not between different vectors. Instead, the real numbers (or more generally, a field) acts to 'scale' vectors. 
So vector spaces are groups, but much more.
You might know that there is something called a field, which is sort of like a group except with two different operations (and some additional conditions). A vector space is like an abelian group that can be acted upon (multiplied by) by the field in a nice, transparent way. There is something more general, called a module, which many people care to study.
A: Every linear space can be viewed as an Abelian group, simply by taking the law of composition in the group as the $+$ operation in the linear space. However, we lose information by doing this.
Here's an example.
Let $G$ denote the Abelian group whose underlying set is $\mathbb{R}^2$ and whose law of composition is the usual sum of vectors. Then the Abelian subgroup of $G$ generated by $\{(1,0)\}$ is
$$\{(n,0) \mid n \in \mathbb{Z}\}.$$
Now on the other hand, let $V$ denote $\mathbb{R}^2$ viewed as a linear space in the usual way. Then the linear subspace of $V$ generated by $\{(1,0)\}$ is $$\{(x,0) \mid x \in \mathbb{R}\}.$$
In the case of $G$, there's really no way (that I can see) of defining the linear subspace generated by $\{(1,0)\}$ so that it agrees with the above set (without looking at the information that we "forgot").
