7
$\begingroup$

Suppose we have a group $\langle S \vert r \rangle \cong F_S /\langle \langle r \rangle \rangle$ where $S$ is a finite set of generators and $r \in F_{S}$, i.e. a finitely generated one realtor group. Suppose $r$ is cyclically reduced (i.e. $r \not = l^c$ for some $\vert l \vert_S < \vert r \vert_S$), then is there some $u \in \langle \langle r \rangle \rangle$ such that $0 < \vert u \vert_S < \vert r \vert_S$?

I believe such a $u$ doesn't exist, however when getting pen and paper out to try and prove it I get a bit stuck (I have tried expressing such a $u$ as a product of conjugated copies of $r$ to play around for a contradiction, as well as a induction on the length of such an expression). I have looked up at the literature surrounding one relator groups and it seems like it is predominantly dealing with the groups themselves not the presentations. I expect the answer to be pretty trivial but I really can't see it, sorry!

The actual question I care about has $u$ as a subword of $r$ (which might make it easier) but I was looking to see if the general question was solved and what methods where involved.

$\endgroup$
9
$\begingroup$

"One-relator theory" can be annoyingly difficult and these "obvious" results tend to be difficult, and almost always use some form of Magnus' method, which you can read about in Combinatorial Group Theory by Lyndon and Schupp (or an other book of the same title by Magnus, Karrass, Solitar). Although maybe it is naive to think it should be easy, after all the only restriction we are placing is that the group can be defined by one relation, why should there be any other connection in the class of one-relator groups?

For the subword of a relation problem, there is a proof in the paper On relators and diagrams for groups with one defining relation by Weinbaum which shows that no subword is trivial in the group (subword proper, not trivial). That means that at least the part of your question you care about holds. Howie also has a proof of this result (and a more general result) in his paper On locally indicable groups. This latter paper gives a more topological interpretation of the Magnus method by using something called towers, which were originally introduced to prove some fundamental results in 3-manifolds, like the loop theorem.

In general there can be $u$ which are trivial in the one relator group, with length less than $r$: Let $r=x^2zxzxz$ and consider $$\begin{align*} r(xz)^{-1}r^{-1} (xz) &=(x^2zxzxz)(xz)^{-1}r^{-1}xz \\ &=(x^2zxz)(x^2zxzxz)^{-1} (xz) \\ &= xz^{-1}x^{-1}z \end{align*}$$ Which has length $4<7$. Similar examples to this can be constructed by finding different basis elements of $F_2$ (which are long with respect to your choosen basis) and use them as relators which give a presentations for $\mathbb{Z}$. That works since you end up with all elements commuting with each other, and you can find a short commutator relation(something similar can be done with $\mathbb{Z}^2$ like my example above). This has the problem that if I changed basis I could make the above relation length $1$, and we would no longer have any strictly smaller words which are relations.

We can come up with a less trivial class of examples though. Consider Baumslag-Solitar groups $BS(m,n)=\langle a,b \mid ba^mb^{-1}=a^n \rangle$, and note that $[a,ba^mb^{-1}]=[a,a^n]=1$, the length of $[a,ba^mb^{-1}]$ is $6+2m$, and the length of the defining relation is $2+m+n$. It is not hard to see that you can arrange $6+2m<2+m+n$ by making $n$ large enough. I am pretty sure that the standard defining relation for $BS(m,n)$ is the shortest defining relation (although I am not one hundred percent positive).

It would be interesting to have a hyperbolic example, where the defining relation is the shortest possible, but there are smaller relations. Random one-relator groups (whatever that means) satisfy small cancellation condition $C'(1/6)$, which implies hyperbolic, and this class of groups do not have relations smaller than the defining relation (by Dehn's algorithm). In fact any Dehn presentation of a group (a group is hyperbolic iff it has a Dehn presentation!) will have the property you will not have relations smaller than the smallest relation in the presentation, by applying Dehn's algorithm.


This is a hyperbolic example update from an answer(and question) of mine on mathoverflow. Below I provide an example of a one ended, one relator, hyperbolic group where there are shorter relators than the defining relation:

Using a handy characterizations in a paper by Ivanov and Schupp called On hyperbolicity of small cancellation groups and one-relator groups.

Consider $\langle a,b,c \mid ab^2ac^{12}\rangle$. By checking Whitehead automorphisms this relation is as short as possible in the $Aut(F_3)$ orbit and has every generator in the relation, so does not have infinitely many ends(Thanks to ADL for the correction). One can look at the abelianization to rule out zero or two ends. Theorem 3 in the above paper says that this group is hyperbolic since it has exactly two occurrences of $a$, no $a^{-1}$, and $b^2c^{-12}$ is not a proper power in the free group generated by $a,b,c$. Now $$(ab^2ac^{12})c(c^{-12}a^{-1}b^{-2}a^{-1})c^{-1}=ab^2aca^{-1}b^{-2}{a^{-1}}c^{-1}$$ which is shorter than the defining relation.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Very nice answer! $\endgroup$ – Moishe Kohan Oct 8 '16 at 0:39
  • $\begingroup$ Thank you Paul, looks like I have some reading to do! I also really like the counter example, that is so counter intuitive. $\endgroup$ – A.Wendland Oct 8 '16 at 8:52
  • $\begingroup$ @A.Wendland I added some more to this answer, in particular a more natural counter example coming from Baumslag-Solitar groups and mentioning that there are a large class of one relator groups where we do not have smaller relations. $\endgroup$ – Paul Plummer Oct 10 '16 at 0:10
  • $\begingroup$ @PaulPlummer Thank you for taking the time to add extra material, the counter example you provided was very useful. I am current studying Plannar Cayley Graphs, and your example is such a graph which has really helped. I know this is a bit more of an advanced question (I really don't know too much about presentations) but is there any general condition which gaurentees that no subword of a relator is a relation? (I think a kinda trivial condition from above would be if all the realtors used disjoint generators, but even this I am starting to question) $\endgroup$ – A.Wendland Oct 15 '16 at 10:35
  • 1
    $\begingroup$ @user1729 I decided to ask the question on MO. $\endgroup$ – Paul Plummer May 17 '18 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.