Given a free group $F$ generated by a set $X$, which by definition is the set of reduced words in $X \cup X^{-1}$, with reduced concatenation of words, I've come across a statement that says every element in a free group is conjugate to a cyclically reduced word, i.e. $\forall g \in F$, $g=k^{-1}hk$ for some $k\in F$ and some cyclically reduced word $h$.
A word $a_1 \cdots a_l$ is cyclically reduced if it is reduced and either $l=0$ or $a_1 \neq a_l^{-1}$.
I cannot come up with a proof of this statement. I would greatly appreciate it if anyone could help me understand this.
If the (reduced) word $w$is cyclically reduced, then there is nothing to prove. If not the first letter of $w$ is the inverse of the last letter: $w=x^{-1}w'x$ where $w'$ is shorter than $w$. Since $w$ is a conjugate of $w'$, we can proceed by induction (a conjugate of a conjugate is a conjugate).