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How can I prove the following result?

Let $G$ be a group, $X\subseteq G$ and let $F_a(X)$ be the free group on $X$. Then the subgroup of $G$ generated by $X$ is isomorphic with $F_a(X)$ if and only if any normal word is distinct from identity $e\in G$.

We call a normal word of $X$ an element $x_1^{r_1}...x_n^{r_n}\in G$ where $x_i\in X$, $x_i\neq x_{i+1}$ and $r_i\neq 0$ are integers.

My real question is, what is the problem if some normal word is identity? I don't see it.

user74411
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    Whats your definition of free group is? For example, a normal word in such a group cannot be the unity since then we'd have a non-trivial relation between the group's generators...but again, this approach can succeed only if it fits your definition (there are at least equivalent basic definitions for free groups that I can think of) – DonAntonio May 13 '13 at 04:39
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    It seems that you can only show that there are no isomorphism $i:\langle X\rangle\to F_a(X)$ such that $i|X=\operatorname{id}$, povided there is a normal word over $X$ distinct from the identity of $G$. In the opposite case the claim seems to be not true. Let $Y$ be a countable infinite set, $G=F_a(Y)$ and $X=Y\cup{y^{-1}}$ for some $y\in Y$. Then $\langle X\rangle$ isomorphic to $F_a(Y)=F_a(X)$, but a normal word $yy^{-1}$ over $X$ is equal to the identity. – Alex Ravsky May 13 '13 at 04:53
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    it seems to be you @AlexRavsky define "free group" on a set $,X,$ as the one with concatenation of formal words with letters from $,X,$ and from $,X^{-1},$ and etc. If this is the case you can not define $,X,$ as you did and form in this way a free group unless you regard $,y^{-1},$ as a very separate, independent symbol (letter) in your alphabet, and if you do so then your element $,y,$ won't possibly have an inverse (or, if it does, then $,y^{-1},$ must already be considered part of the elegible letters...) – DonAntonio May 13 '13 at 05:20
  • @DonAntonio I mainly agree with you. It is a misprint in my answer, should be “$F_a(Y)$, and the free groups $F_a(Y)$ and $F_a(X)$ are isomorphic” instead of “$F_a(Y)=F_a(X)$”. The element $y^{-1}$ of $F_a(X)$ has no inverse in $X$. Nevertheless, the normal word $yy^{−1}$ of the group $\langle X\rangle\subset G$ is equal to the identity. – Alex Ravsky May 13 '13 at 18:52
  • Oh, ok then.... – DonAntonio May 13 '13 at 22:55
  • Warning: according to https://en.wikipedia.org/wiki/Normal_form_for_free_groups_and_free_product_of_groups#Definition_.28normal_form.29 your definition of normal word is different, being Lyndon et al the reference – janmarqz Jan 06 '14 at 23:04

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