I'm going through the proof of the following:
If $F$ is a free group, and $1 \neq w \in F$, then $|w| = \infty$, i.e. $F$ is torsion-free.
In the proof, we let $F=F_X$ be the free group on $X$, so $w$ is a nonempty reduced word. The proof goes on to say that, in general, we can write $$w = a_1 \cdots a_r a_{r+1} \cdots a_{l-r} a_r^{-1}\cdots a_1^{-1},$$ with $a_{r+1} \neq a_{l-r}^{-1}$, i.e. $w = \beta\alpha\beta^{-1}$, with $\alpha$ a nonempty cyclically reduced word.
Why is it possible to write $w$ in such a way?
This is really trivial when you see what it means. Consider the unique representation of $w$ as a reduced word, say $$w=a_1a_2\dots a_n.$$ If $a_1\neq a_n^{-1}$, then we can take $r=0$. Otherwise, if $a_2\neq a_{n-1}^{-1}$, we can take $r=1$. Otherwise, if $a_3\neq a_{n-2}^{-1}$, we can take $r=2$. And so on. Assuming $n>0$, we must eventually reach an $r$ such that $a_{r+1}\neq a_{n-r}^{-1}$, since eventually we reach the middle of the word and the left and right halves of the word would cancel out, contradicting the assumption that $w$ is reduced.
That is, $\beta$ is just the longest initial subword of $w$ whose inverse appears at the end of $w$, and $r$ is the length of that subword.