String powers. Lemma 1.3.5 in word processing in groups. Epstein.

Question

I can't understand this demonstration. Why if ${w'}_1$ is different from ´${w'}_2$ then we have that $f(u)$ and $v'$ are powers of some string $z$?

Answer 1

The way induction is used is not described in much detail. I try to understand and explain it. The base case where $|f(u)|+|f(v)| = 0$ is not even mentioned.

For the induction step, instead of the alphabet $\{u,v\}$ the alphabet $\{u,v'\}$ is used, supposedly with the map $f'(u) = f(u); f'(v') = v'$. Since $f(u)v' = f(v)$ by the definition of $v'$, $|f'(u)| + |f'(v')| = |f(v)|$. Unless $f(u)$ is empty, $|f'(u)| + |f'(v')| < |f(u)|+|f(v)|$ and therefore the induction hypothesis applies to $f(u)$ and $v'$. If $f(u)$ is empty it is a common power of any string.