# 1-related groups

Every 1-related group $G$ with at least 2 generators is an HNN extension of another 1-related group $G_1$ with free associated subgroups. Indeed, if the total exponent of one letter in the relator is 0, then one can take that letter as the free letter. If there are two letters $a$ and $b$ in the relator $r$ and $a$ occurs with the total exponent $m$, $b$ occurs with total exponent $n$, and, say $m\ge n>0$ (the other cases are similar), then make a substitution $a=a$, $b=b'a^{-1}, c=c,...$ (i.e. change the generating set accordingly). In the resulting 1-related presentation $\langle a',b,c,...| r(a,b'a^{-1},c...)\rangle$ the total exponent of $a$ is $m - n < m$, and the total exponents of other letters stay the same. By iterating this procedure, one can reduce the general case to the case of total exponent 0. Note that the relator of $G_1$ can be longer than the relator of $G$. I think it is almost obvious then that starting with a cyclic group, and using HNN extensions with free associated subgroups, one can get any 1-related group (and all intermediate groups will be 1-related also).

Question. Given a 1-related group $G$, what is the shortest representation of $G$ as a sequence of HNN extensions as above? Can one estimate that length from above in terms of the length of the relator?

Update I always thought that the length of the sequence should be linear in $|r|$. Ian Agol's answer below suggests a quadratic upper bound. Is there a linear upper bound?

-
I think you can get a lower bound on the length from below in terms of the depth in the derived series, but I expect this is far from a sharp lower bound. –  Ian Agol May 30 '11 at 1:42
Probably even the depth in the lower central series would be a lower bound. But isn't the depth linear in length? As I understand a similar problem exists for Heegard splittings. What is the lower bound there? –  Mark Sapir May 30 '11 at 2:50