0

Let $F=<X>$ be a free group and $v, w \in F$ such that $v^n \cdot w^m = w^m \cdot v^n$. Show that $v\cdot w = w\cdot v$.

I didn't think it'd be very hard, but it turns out I'm stuck. I've thought of using induction on the lengths of $v$ and $w$, or induction on $n$ and $m$ but I didn't manage to get anything out of it. So if the solution includes induction and you don't want to give it out immediately but with a hint, please be a little more specific than just suggesting to use induction. Thanks in advance for your help.

Nothing just that
  • 153
  • Using the (normal or not) reduced form for words in a free group, we get that $;v^nw^m=w^mv^n;$ is possible iff $;u,v;$ are powers of the same element in $;F;$ . This is due that in a free group there exist no non-trivial relations between elements of a free basis. – DonAntonio Jan 26 '21 at 23:49
  • I don't think I understand why this relation being trivial implies that $v$ and $w$ are powers of the same element. At least it's not immediate at all. In their reduced forms, there may be many cancelations. I've tried working out the different cases but it got me nowhere. Is there a more direct way? – Nothing just that Jan 27 '21 at 10:44
  • I can't see any other way. Perhaps induction on the sum of the lenghts of $;v,,w;$ as reduced words...and first of all to deduce what happens with $;v^nw^m=w^mv^n;$ in this form – DonAntonio Jan 27 '21 at 18:34

0 Answers0