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Is my proof for the fact that every nonidentity element in a free group has infinite order correct?

Suppose that $F$ is the free group on a nonempty set $X$. We have $F\simeq \prod_{x\in X}^w <x> \simeq \sum_{x\in X}\mathbb Z$, and their torsion subgroups are isomorphic too, but the torsion subgroup of $\sum_{x\in X}\mathbb Z$ is the the trivial subgroup so the proof is complete.

Appendix: after the answer, I realized this proof is only applicable for the abelian case, not for the general case

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  • $\begingroup$ If you mean free abelian group, then I can believe it. The theorem is true for free group also, but you need a different proof. $\endgroup$ – Somos May 29 '18 at 23:57
  • $\begingroup$ Yes, I’m new to this topic and unknowingly my proof was true only for the abelian case $\endgroup$ – user555729 May 30 '18 at 9:40
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A free group is not generally isomorphic to a direct product of cyclic groups; e.g., the free group on two elements is not isomorphic to $\mathbb Z \times \mathbb Z$ (for one, it's not abelian).

Instead, think directly about what it would mean for an element to have finite order.

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