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?

  • $\begingroup$ 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. $\endgroup$
    – Ian Agol
    May 30 '11 at 1:42
  • $\begingroup$ 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? $\endgroup$
    – user6976
    May 30 '11 at 2:50

Check out the proof of Theorem 4.1 of Joe Masters' paper. Given a 1-relator group presentation, realize the free group as the fundamental group of a compact surface (with boundary), and the relator as an immersed loop in this surface. Choose such a surface so that the self-intersection number of the loop representing the relator is minimal. Then Masters' proof shows that this self-intersection number decreases after each (non-trivial) HNN extension in the Magnus-Moldavansky hierarchy (for the base case of an embedded loop on a surface, notice that the 1-relator group is either free or a surface group if the loop is parallel to the only boundary component). I think one can at least bound this self-intersection number from above quadratically in the length of the relator, so this gives a crude estimate.

  • $\begingroup$ Perhaps it's worth mentioning that this proof does not need to be carried out on a punctured surface -- you can also use cyclic tower maps of graphs. Jim Howie proved that such a tower is finite, and again, there's an upper bound deriving from the fact that the number of self-intersections decreases at each stage. In fairness, I suppose you should get a better bound from thinking of it on a surface, since there will be fewer self-intersections, but it will be more complicated to read off the word. $\endgroup$
    – HJRW
    Oct 29 '20 at 9:29

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