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Theorem: Suppose that $F$ is a free group on a nonempty set $X$ (with $i$ as the map between them). Prove that every nonidentity element has infinite order.

Proof: We identify $X$ with its image under $i$. We have $F\cong \prod_{x\in X}^*<x> \cong \prod_{x\in X}^*\mathbb Z$ (where $\prod^*$ is the notation for free product). They have isomorphic torsion subgroups. The torsion subgroup of $\prod_{x\in X}^*\mathbb Z$ is the trivial subgroup so the proof is complete.

(The proofs in Every nonidentity element in a free group $F$ has infinite order did not appeal to me so I wanted to provide a short elegant proof. In addition, the proof here is the extension of the method used in A short proof that every nonidentity element in a free group has infinite order.)

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  • $\begingroup$ ... and you claim that the formula for the torsion group of a free product is easier than the result we are proving? And not merely the same thing in fancier language? $\endgroup$ – GEdgar May 30 '18 at 11:27
  • $\begingroup$ The formula for the torsion subgroup of the $\color{blue} {particular} $ $\prod^*\mathbb Z$ is obvious @GEdgar $\endgroup$ – user555729 May 30 '18 at 11:33
  • $\begingroup$ That "obvious" thing is exactly what the question asks, right? So I'm saying, your proof is circular. $\endgroup$ – GEdgar May 30 '18 at 11:44
  • $\begingroup$ I am only a toddler in mathematics, but I think algebra started from the integers $\mathbb Z$ so it is natural that many good “obvious” algebraic things happen in $\mathbb Z$, so if one can reduce their algebraic problem to $\mathbb Z$ their life will be good @GEdgar $\endgroup$ – user555729 May 30 '18 at 11:49
  • $\begingroup$ @GEdgar now I see why you considered that as circular. By a free group I did not necessarily mean the set of words. $\endgroup$ – user555729 May 30 '18 at 12:33
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The proof works, but I agree (in part) with GEdgar in the comments that it's not immediately obvious what the torsion subgroup of the free product of copies of $\mathbb Z$ is. Perhaps you should justify your claim that it is trivial.

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  • $\begingroup$ @1319m Perhaps you can explain it to me, then. $\endgroup$ – BallBoy May 30 '18 at 12:04
  • $\begingroup$ @1319m If you say so. I can't claim to speak for algebraists. To me, this fact and the fact you set out to prove are at an approximately equal level of triviality. $\endgroup$ – BallBoy May 30 '18 at 12:54
  • $\begingroup$ They are at an equal level of triviality if you work with the set of reduced words on $X$, but $F$ can be any other free group. But since the set of reduced words is usually the only familiar free group I agree with you and GEdgar $\endgroup$ – user555729 May 30 '18 at 12:59

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