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I take as a definition of free group on a set S as the codomain F of the universal element f in the category of functions g from S to group G, where morphism from g to h are function j from the group H to the group G such that j•h=g. I was wondering about how to use the universal properties to prove the every element in Rg(f) has infinite period. Any help is really appreciate

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There is a way to prove this without using reduced words, but it is much harder than necessary and probably goes beyond what you know (to the point that this proof becomes just ridiculous). However, here it is (since you asked):

Step 1. Use the universal property of $F=F(S)$ to construct a fixed-point free action of $F$ on a tree $T$ (a connected graph without circuits). One can prove this by, say, appealing to Seifert-van Kampen theorem applied to the wedge $V$ of $|S|$ circles, as it is done in Massey's book on algebraic topology and then take the universal cover $T$ of $V$. Serre also does this, if I remember it correctly, in his book "Trees".

Step 2. Prove Cartan's fixed point theorem for trees: Every finite group of automorphisms of a tree fixes a point. (Serre does it as well, I think: It is a nice combinatorial argument.)

Setp 3. Conclude that if $F(S)$ has a nontrivial finite order element, it has to fix a point in $T$, which is a contradiction.

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  • $\begingroup$ Thx you for your answer, but just like you said it is a bit too much for me. $\endgroup$ Aug 24, 2017 at 7:08

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