Free splittings of one-relator groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T16:14:17Z http://mathoverflow.net/feeds/question/26640 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26640/free-splittings-of-one-relator-groups Free splittings of one-relator groups HW 2010-06-01T00:27:21Z 2010-06-28T22:28:02Z <p>Roughly speaking, I want to know whether one-relator groups only have 'obvious' free splittings.</p> <blockquote> <p>Consider a one-relator group $G=F/\langle\langle r\rangle\rangle$, where $F$ is a free group. Is it true that $F$ splits non-trivially as a free product $A * B$ if and only if $r$ is contained in a proper free factor of $F$?</p> </blockquote> <p><strong>Remarks</strong></p> <ol> <li>One direction is obvious. It is clear that if $r$ is contained in a proper free factor then $G$ splits freely. (We think of $\mathbb{Z}\cong\langle a,b\rangle/\langle\langle b\rangle\rangle$ as an HNN extension of the trivial group, so it's not really a counterexample, even though it might look like one.)</li> <li>A quick search of the literature suggests that the isomorphism problem for one-relator groups is wide open. (I'd be interested in any details that anyone may have.)</li> <li>There is no decision-theoretic obstruction. Magnus famously solved the word problem for one-relator groups. Much more recently, <a href="http://arxiv.org/abs/0906.3902" rel="nofollow">Nicholas Touikan has shown</a> that, for any finitely generated group, if you can solve the word problem then you can compute the Grushko decomposition. So one can algorithmically determine whether a given one-relator group splits. If the answer to my question is 'yes' then one can use Whitehead's Algorithm to find this out comparatively quickly.</li> <li>When I first considered this question, it seemed to me that the answer was obviously 'yes' - I don't see how there could possibly be room in a presentation 2-complex for a 'non-obvious' free splitting. But a proof has eluded me, and of course many seemingly obvious facts about one-relator groups are extremely hard to prove.</li> </ol> http://mathoverflow.net/questions/26640/free-splittings-of-one-relator-groups/26660#26660 Answer by Agol for Free splittings of one-relator groups Agol 2010-06-01T03:53:47Z 2010-06-01T03:53:47Z <p>Your question is reminiscent of <a href="http://www.ams.org/mathscinet-getitem?mr=754723" rel="nofollow">Jaco's lemma</a>. A special case of Jaco's lemma applies to a 2-handle attached to the boundary of a (3-dimensional) handlebody $H$ (which has free fundamental group). If the boundary curve $J\subset \partial H$ along which the handle is attached is "disk-busting", that is, $\partial H-J$ is incompressible (and therefore $\pi_1$-injective) in $H$, then the manifold obtained has $\pi_1$-injective boundary (and therefore does not split as a free product). This condition is easily seen to be equivalent to the conjugacy class of $J$ not belonging to any free factor of $\pi_1(H)$. So this answers your question in this very special case (also note that this works for "orbifold" handles attached along $J$). It's not clear whether Jaco's method might apply in your case, but it might be worth having a look (there are other proofs and generalizations of it too which you can find through Mathscinet). In particular, his argument might also apply if the word is only virtually geometric. </p> http://mathoverflow.net/questions/26640/free-splittings-of-one-relator-groups/26673#26673 Answer by Tom Church for Free splittings of one-relator groups Tom Church 2010-06-01T05:49:38Z 2010-06-01T05:49:38Z <p>I'm not sure whether it helps here, but your question reminds me of the Freiheitssatz. As you probably know, this is the theorem of Magnus that says that if $G=\langle x_1,\ldots,x_n|r\rangle$ and $r$ involves the generator $x_n$, then the elements $x_1,\ldots,x_{n-1}$ generate a free group of rank $n-1$ inside $G$. Certainly your assumption implies this hypothesis with respect to any basis $x_1,\ldots,x_n$.</p> <p>Also, I feel like we don't want to count HNN extensions as free products -- can't we just exclude the example of $\mathbb{Z}=\langle a,b|b\rangle$ by fiat? There aren't any other such counterexamples, right?</p> http://mathoverflow.net/questions/26640/free-splittings-of-one-relator-groups/26767#26767 Answer by Agol for Free splittings of one-relator groups Agol 2010-06-01T23:28:00Z 2010-06-01T23:28:00Z <p>I think <a href="http://en.wikipedia.org/wiki/Grushko_theorem" rel="nofollow">Grushko</a> plus the <a href="http://en.wikipedia.org/wiki/Freiheitssatz" rel="nofollow">Freiheitssa</a>tz does the trick. Suppose that $G=A\ast B$ is a 1-relator group which splits as a free product non-trivially. By Grushko, $rank(G)=rank(A)+rank(B)=m+n$, where $rank(A)=m, rank(B)=n$. If $G$ is not free, then by Grushko there is a 1-relator presentation $\langle x_1,\ldots, x_m ,y_1,\ldots, y_n | R\rangle$, such that $\langle x_1,\ldots, x_m \rangle =A \leq G, \langle y_1, \ldots, y_n \rangle=B \leq G$ (for this, one has to use the strong version of Grushko that any 1-relator presentation is Nielsen equivalent to one of this type). Suppose that $R$ is cyclically reduced, and involves a generator $x_1 \in F_m$. Then $\langle x_2,\ldots , x_m, y_1, \ldots, y_n\rangle$ generates a free subgroup of $G$ by the Freiheitssatz. But this implies that $B = \langle y_, \ldots, y_n\rangle$ is free. Moreover, if $G$ involves one of the generators $y_i$, then one sees that $A=\langle x_1,\ldots,x_m\rangle$ is also free, and therefore $G=A\ast B$ is free, a contradiction. So $R \in F_m$, as required.</p> http://mathoverflow.net/questions/26640/free-splittings-of-one-relator-groups/29839#29839 Answer by Diane for Free splittings of one-relator groups Diane 2010-06-28T22:28:02Z 2010-06-28T22:28:02Z <p>I just came across the following, which is Prop. II.5.13 of Lyndon-Schupp.</p> <p>Proposition. Let $G = \langle x_1, \ldots , x_n: r\rangle$ where $r$ is of minimal length under $Aut(\langle x_1, \ldots , x_n\rangle)$ and contains exactly the generators $x_1,\ldots , x_k$ for some $0 \leq k \leq n$. Then $G \cong G_1*G_2$ where $G_1 = \langle x_1, \ldots , x_k:r\rangle$ is freely indecomposable and $G_2$ is free with basis $x_{k+1}, \ldots ,x_n$.</p> <p>Unless I'm missing something, up to isomorphism you can assume that your relator has minimal length. If it is not contained in a free factor of $G$, then $k=n$ in the Proposition, hence $G = G_1$ is freely indecomposable.</p>