Q:

free abelian group

let $X=\{a_i\mid i\in I\}$ be a set , then the free abelian group on X is (isomorphic to) the group defined by the generators X and the relations (in multiplicative notation) $\{a_ia_ja_i^{-1}a_j^{-1}=e\mid i,j\in I\}$
here is where i need help..
let the free group on X be F , let N be the normal subgroup of F that is generated by the relations set Y , so i need to prove $G=F/N$ is abelian group and it has a nonempty basis .
to prove the first one is easy , i am stuck on how to prove $F/N$ has a nonempty basis because i don't know what the element in group N general is 

A:

From my point of view this is trivial but I could be missing something important from the OP's view.
First, $\;F/N\;$ abelian as clearly $\;F'\le N\;$ by definition of $\;N\;$ . About the non-empty set of generators the reason why I think this is trivial is that if $\;F=\langle\;f_i\;:\;i\in J\;\rangle\;$, then
$$\langle\;f_iN\;:\;i\in J\;\rangle=F/N$$
The only way the above can be empty is if $\;J=\emptyset\iff F=\{1\}\;$ , but for this trivial case...
Also, even if $\;F\;$ is abelian the above cannot be empty, but also this case can be considered trivial and we can, probably, assume $\;|J|\ge 2\;$

