free abelian groups and abelian groups that are free What is the difference between a "free abelian group" and a free group that is abelian or an abelian group that is free.. it seems very confusing because in my understanding a free group $G$ is a group that has a basis $B=(b_i)_i$, that is $\forall g\in G$, $g=\prod{b_i^{m_i}}$ ,$m_i\in \mathbb Z$. In particular when $G$ is abelian, we write this additively and we have $g=\sum_{i}{m_ib_i}$, $m_i\in\mathbb Z$ and we call this a "free abelian group" is this correct?
 A: Your error is in describing the (absolutely) free group.
A free group with "basis" $\{b_i\}$ does not consist of elements of the form $\prod b_i^{m_i}$. 
A free group with "basis" $\{b_i\}_{i\in I}$ consists of all finite sequences of the form
$$(b_{i_1}^{\epsilon_1},\ldots,b_{i_m}^{\epsilon_m})$$
where $m\geq 0$, $\epsilon_j\in\{1,-1\}$ for each $j$, $i_j\in I$ for each $j$; and there does not exist $j$ such that $i_j=i_{j+1}$ and $\epsilon_j=-\epsilon_{j+1}$. These are called "reduced words".  The set $\{b_i\}_{i\in I}$ is not usually called a basis, but rather a "free generating set".
The multiplication rule is defined as by "juxtapose and cancel": if 
$$(b_{i_1}^{\epsilon_1},\ldots,b_{i_m}^{\epsilon_m})$$
and
$$(c_{k_1}^{\delta_1},\ldots,c_{i_n}^{\delta_n})$$
are two reduced words, then their product is defined recursively to be:
$$\begin{align*}
(b_{i_1}^{\epsilon_1},\ldots,b_{i_m}^{\epsilon_m})&*(c_{k_1}^{\delta_1},\ldots,c_{k_m}^{\delta_n})\\ 
&= \left\{\begin{array}{ll}
(b_{i_1}^{\epsilon_1},\ldots,b_{i_m}^{\epsilon_m}) &\text{if }n=0;\\
\\
(c_{k_1}^{\delta_1},\ldots,c_{k_n}^{\delta_n}) &\text{if }m=0;\\
\\
(b_{i_1}^{\epsilon_1},\ldots,b_{i_m}^{\epsilon_m},c_{k_1}^{\delta_1},\ldots,c_{k_n}^{\delta_n}) &\text{if }i_m\neq k_1\text{ or }(i_m=k_1\text{ and }\epsilon_m=\delta_1)\\
\\
(b_{i_1}^{\epsilon_1},\ldots,b_{i_{m-1}}^{\epsilon_{m-1}})*(c_{k_2}^{\delta_2},\ldots,c_{k_n}^{\delta_n})&\text{if }i_m=k_1\text{ and }\epsilon_m = -\delta_1.
\end{array}\right.
\end{align*} $$
The idea is that the sequence corresponds to simply an expression of products of $b_i$ and their inverses, in which the only things we are allowed to "simplify" are $b_ib_i^{-1}$ and $b_i^{-1}b_i$. Everything else is "left indicated".
Free groups are difficult objects to get used to when you first encounter them. But there are lots and lots of way of thinking about them: formally, as above, and playing directly with the words and elements. "Universally", via the universal property; you can find three different ways of thinking about them in that context in George Bergman's Invitation to General Algebra and Universal Constructions (see Chapter 2 for a great discussion of free groups). You can think of them "topologically", as fundamental groups of graphs. And other ways. 
What you cannot do in a free group is freely permute basis elements. In a free group, $b_ib_j = b_jb_i$ if and only if $i=j$. So you cannot express arbitrary elements as you claimed.
A: Although it is useful to think of free groups in the way you describe them, that isn't really what they are. The free-ness is to do with the fact that they are the free objects in the category of groups. Similarly, the free abelian groups are the free objects in the category of abelian groups. It just so happens that the definitions you give (which are correct, in an informal way, so long as you point out they such a word is never the trivial word unless it is freely trivial) coincide with these definitions.
Basically, given a function from $S$ to a group $G$ there exists a unique homomorphism $\varphi: F_S\rightarrow S$ such that the following diagram commutes,

where $F_S$ is the free group on the set $S$ (this $S$ is your basis $B$). Now, given a function $S$ to an abelian group $G$ there exists a unique homomorphism $\varphi: A_S\rightarrow S$ such that the above diagram commutes (if we replaced $F_S$ with $A_S$), where $A_S$ is the free abelian group on the set $S$.
It is a relatively easy exercise (or at least, the proof can be found in most decent group theory texts) to prove that if $|S|\geq 2$ then $F_S$ does not commute, and I will leave you to prove it. However, if $S=1$ then $F_S$ commutes (why?). So, the only commutative free group is $F_{a}(\cong \mathbb{Z})$.
A: This is what i understand right now:
A basis is defined for only an abelian group. An abelian group $G$ has a basis $B=(b_i)_i$, if $\forall g\in G$, there exists unique $(m_i)_i$ integers $\in \mathbb Z$ such that $g=\sum{m_ib_i}$. This definition of basis can not be generalized to a non abelian group since associating an $m_i$ for each $g_i$ is no longer possible as $b_i$ can be found in may locations, for example suppose a basis is $B=\{b_1,b_2,b_3\}$ then an element like $b_1^2b_2^5b_1^{-1}b_3^4b_2^3$ can not be described by a integers $(2,5,-1,4,3)$. That's why for a general group we don't talk about a basis but we talk about a set $B$ that freely generates $G$ in the sense that each $g$ is uniquely written as a  product of elements of $B$ and its inverses. 
