Prove every free group is torsion-free.
Here is my attempt:
Let $F$ be a free group and let $a \in F$ be an element of finite order, i.e. $a^n=1$ for some $n > 0$. Also $a=a_1…a_s$ is a reduced word of length $s>0$, i.e. $a$ is an non-identity element.
It suffices to show a contradiction. But I don't know exactly how.
Best Answer
Hint: given a word $a \not = e$ (not assumed to be torsion! we're proving this directly, not by contradiction), write it as $u w u^{-1}$ with no cancellation between $u$ and $w$, and no cancellation between $w$ and $u^{-1}$, where $w$ is cyclically reduced: it is a reduced word $a_0 a_1 \dots a_m$ such that $a_0 a_m \not = e$.
Now consider powers of the word: $a^n = u w^n u^{-1}$.
There is no cancellation between $u$ and $w^n$; and there is no cancellation between $w^n$ and $u^{-1}$; and there is no cancellation within $w^n$ because we defined $w$ to be cyclically reduced. So $a^n$ is not the identity word, because its length is not $0$.