# Can a hyperbolic, one ended, one relator group, have a shorter trivial word?

Let $G= \langle S \mid r \rangle$ be a one-relator presentation for a one-ended hyperbolic group, with $r$ cyclically reduced.

Question: Can there be a nontrivial word $w(S)$ which is trivial in the group $G$ but has length shorter than $r$? What if $r$ is the shortest possible word for a one-relator presentation of $G$?

Note that it follows from Newman's spelling theorem that in the torsion case there are no shorter words, since you can apply Dehn's algorithm. Similarly if $r$ gives a $C'(1/6)$ presentation there are no shorter words.

Generally it is known that subwords of $r$ will not be trivial either. This is proved by Weinbaum in On relators and diagrams for groups with one defining relation.

This question grew out of this question on math.se and my answer to it. One thing to note is that without hyperbolicity you can find that some Baumslaug-Solitar groups provide examples with shorter trivial words.

• Just for context, in the presentation $\langle a,b,\dots|ab^k\rangle$, the word $[ab^k,b]=ab^kbb^{-k}a^{-1}b^{-1}=[a,b]$ of length 4 is a relation, so the relator is not shortest when $k\ge 4$. This does not answer the question since this group is free, hence hyperbolic but not 1-ended. – YCor May 18 '18 at 22:38

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.
• Nice example! I just wanted to point out that your group is one-ended, but not for the reasons you state. For example, the following group is not one-ended even though the relator is not primitive and every generator occurs in the relator: $\langle a,b,c,d\mid (ad)b^2(ad)c^{12}\rangle$. The theorem you want is as follows: Suppose $R$ has shortest possible length in its $Aut(F(\mathbf{x}))$-orbit. Then $G=\langle \mathbf{x}\mid R\rangle$ is infinitely-ended if and only if there exists a generator which does not occur in $R$. – ADL May 22 '18 at 11:47
• (If $|\mathbf{x}|=2$ and $R$ is primitive then $G$ is two-ended, otherwise $G$ is one-ended. Note that your relator satisfies this theorem, which can be seen by using the Whitehead automorphisms to verify minimality of its length. So $G$ is not infinitely-ended, and hence has to be one-ended (erm......there is presumably an easy way of showing non-0 and non-2 endedness.....but for your group you can use the Baumslag-Pride result to see that it is large and so not 0- or 2-ended.).) – ADL May 22 '18 at 11:57
• @ADL, a tiny comment: the theorem you refer to actually checks more-than-one-endedness, doesn’t it? Consider, for instance, $\langle x,y\mid y\rangle$. – HJRW Jun 10 '18 at 7:22
• @HJRW Yes, my mistake. It should read "... Then $G$ has more than one end if and only if there exists a generator which does not occur in $R$." The first sentence in my second comment ("If $|\mathbf{x}|=2$ and...") is also incorrect. More can be said regarding this test for endedness, but I guess here isn't the place! – ADL Aug 24 '18 at 12:51