TITLE: free abelian groups and abelian groups that are free
QUESTION [7 upvotes]: 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?

REPLY [5 votes]: 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.