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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

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    $\begingroup$ What is your question? $\endgroup$ – Tobias Kildetoft Dec 5 '14 at 14:19
  • $\begingroup$ prove this fact :) $\endgroup$ – Idele Dec 5 '14 at 14:20
  • $\begingroup$ So what did you try? Please show your work and tell us where you got stuck. $\endgroup$ – Harald Hanche-Olsen Dec 5 '14 at 14:22
  • $\begingroup$ And what definition of the free abelian group are you using? $\endgroup$ – Tobias Kildetoft Dec 5 '14 at 14:22
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    $\begingroup$ Have you tried to define a map in one direction or the other? There are not that many logical choices. $\endgroup$ – Tobias Kildetoft Dec 5 '14 at 14:33
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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 $\;|J|\ge 2\;$

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  • $\begingroup$ sorry i didn't make it clear . an abelian group G's nonempty basis X is a nonempty set that satisfy: <X>=G , and if $n_1x_1+n_2x_2+\cdots+n_rx_r=0$ then $n_1=n_2=\cdots=n_r=0$,where $x_k\in X,k=1,2,\cdots,r,n_k \in Z$ $\endgroup$ – Idele Dec 6 '14 at 15:52
  • $\begingroup$ @hctb, that's exactly what we have here in my answer. $\endgroup$ – Timbuc Dec 6 '14 at 18:45
  • $\begingroup$ but how to prove if $k_1,k_2,\cdots,k_r\in J,n_k\in Z$ and $f_{k_1}^{n_1}f_{k_2}^{n_2}\cdots f_{k_r}^{n_r}N=N$ which is $f_{k_1}^{n_1}f_{k_2}^{n_2}\cdots f_{k_r}^{n_r}\in N$ then $n_1=n_2=\cdots=n_r=0$ $\endgroup$ – Idele Dec 7 '14 at 2:24
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    $\begingroup$ @hctb Using for example the canonical normal form for an element in a free group. Try to prove the following very nice, and for me once surprising, theorem: if $\;F:=F(a_1,...,a_m)\;$ is the free group on $\;\{a_1,...,a_m\}\;$ and $\;w:=w(a_1,...,a_m)\;$ is a word in these letters, then the exponent sum $\;e_i\;$ of a letter $\;a_i\;$ is the sum of all the exponents to which the letter $\;a_i\;$ is raised in the word. For example, $\;e_2(a_1^2a_5a_6a_2^{-4}a_3a_2^2)=-2\;$ . The theorem is: a letter $\;w\;$ belongs to $\;F':=[F:F]\;$ iff $\;e_i=0\;$ for all $\;i\;$ . $\endgroup$ – Timbuc Dec 7 '14 at 3:17

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