1

I'm really struggling with the inductive proof of the associativity of free groups, given about halfway down page 6 of this pdf.

The bit I'm not getting is this:

Suppose now that bc involves a cancellation; so c = b-1v starts with b-1. Therefore, (ab)c = wc = w(b-1v) while a(bc) = a(v) = (wb-1)v. Since l(w) + l(v) < l(a) + l(c), the last two terms are equal by induction hypothesis. Hence, the associativity holds when l(b) = 1.

I don't understand what $l(a) + l(c)$ has to do with anything, and why the fact that $l(w) + l(v) < l(a) + l(c)$ means that "the last two terms are equal". I am familiar with proof by induction, but I don't get how it works in this case.

Any help would be greatly appreciated!

A K Lemming
  • 11
  • It looks as though the proof is by induction on $l(a)+l(c)$. The base case is when $l(a)=l(c)=0$, and the claim is trivial. Since $l(w)+l(v)<l(a)+l(c)$, we can assume by induction that the result is true when $a,b,c$ are replaced by $w,b^{-1},v$. Hence $w(b^{-1}v)=(wb^{-1})v$. – Derek Holt Nov 16 '15 at 13:20
  • 1
    There are ways to construct free groups without the fuzz of equivalence classes, reduced words, struggle with associativity, etc. If you are interested then have a look here. – drhab Nov 16 '15 at 13:55
  • @DerekHolt Oh I think I get it now -- I wasn't familiar with strong induction as opposed to weak induction. We know associativity holds for the base case $l(a) + l(c) = 0$, and if we assume that it holds for $l(w) + l(v) < l(a) + l(c)$ then it follows that it also holds for $l(a) + l(c).$ Therefore, it holds for all $l(a) + l(c)$. – A K Lemming Nov 16 '15 at 13:59

0 Answers0