Every nonempty reduced word has the form $tc{t}^{-1}$ with $c$ nonempty and cyclically reduced

A reduced word of length one is nonempty and cyclically reduced, so the assertion holds with $t=\epsilon$ and $c=w$.

Assume the assertion for all nonempty reduced words shorter than $w$. If $w$ is cyclically reduced, take $t=\epsilon$ and $c=w$.

If $w$ is not cyclically reduced, [F1] says that its first and last letters are inverse, so $w=au{a}^{-1}$ literally; reducedness of $w$ makes $u$ nonempty and reduced, and $\mathrm{\mid }u\mathrm{\mid }=\mathrm{\mid }w\mathrm{\mid }-2$.

Apply [L1] to the property that the assertion holds at every length at most $n$. The induction hypothesis then applies to the shorter word $u$, so write $u=t^{\mathrm{\prime }}c(t^{\mathrm{\prime }}{)}^{-1}$ with $c$ nonempty and cyclically reduced; then $w=(a{t}^{\mathrm{\prime }})c(a{t}^{\mathrm{\prime }}{)}^{-1}$ literally.

The alternatives in steps 1.2 and 2.1 cover every nonempty reduced word and give the required factorisation, including the one-letter boundary.

In the group of [L2] the product of reduced words is their concatenation followed by free reduction. The concatenation $tc{t}^{-1}$ is the reduced word $w$ of step 3.1, so no reduction occurs there and that product is $w$; the concatenation $t{t}^{-1}$ reduces to the empty word, which is the identity because concatenating it with any reduced word changes nothing, so $t^{-1}$ is the inverse of $t$. Hence $w=tc{t}^{-1}$ exhibits $w$ as a conjugate of $c$ in that group.