My definitions:
A free abelian group is an abelian group which has a basis.
(Here, a basis of a free abelian group is a subset of the group which generates it and they are $\mathbb{Z}$-independent)
My question:
A free abelian group is not free unless it is cyclic but we still call it 'free' because, i guess,
a free abelian group $G$ has only relation $ab=ba$ for all $a,b \in G$.
To justify my guess, i've tried to prove the following:
Let $F$ be a group on a set $X$ and $F'=\langle aba^{-1}b^{-1} \mid a,b \in F \rangle$ be the commutator group of $F$.
Then the quotient group ${F}/{F'}$ is a free abelian group so it has a basis.
In this case, how can i find a basis of $F/F'$?