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A free group contains no notrivial elements of finite order

This statement seems obvious and trivial, but I cannot think of a nice proof besides going and getting my hands dirty with elements of the group. Is there a nice proof of this fact?

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Depends how much you know about free groups. If you know that subgroups of free groups are free, then this is easy (since the existence of groups that have elements that are not torsion shows the free group of rank 1 is infinite cyclic). If the element is not a commutator word (nontrivial image in the free abelian group), then simply map to cyclic group by mapping an appropriate letter to the generator and the rest to $e$. Otherwise, I'd say you do have to do a bit of mucking about with elements, but not much: prove it for cyclically reduced elements, and you are done. –  Arturo Magidin Nov 6 '11 at 3:14

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Assume $F$ is the free group on a set $X$ and $a \in F$ has finite order $e$. Since $F(X)$ is the directed colimit of the $F(Y)$ with $Y \subseteq X$ with $Y$ finite, we may assume as well that $X$ is finite. Since every element is conjugated to a cyclically reduced word, we may assume that $a=x_{k_1}^{e_1} \dotsc x_{k_n}^{e_n}$ is cyclically reduced with $e_i \neq 0$; in particular $x_{k_1} \neq x_{k_n}$. Then $a^e=x_{k_1}^{e_1} \dotsc x_{k_n}^{e_n} x_{k_1}^{e_1} \dotsc x_{k_n}^{e_n} \dotsc $ is again (cyclically) reduced. On the other hand it is trivial, thus it's length must be $0$. This shows $a=1$.

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Thanks, this was pretty much what I ended up doing. I guess the proof that subgroups of free groups are free is the prettiest way to do this, but you pay the price of having to go and prove that theorem. On the other hand, that theorem has some nice proofs, including a beautiful topological one. –  EWEBEWE Nov 6 '11 at 11:59

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