Free splittings of one-relator groups Roughly speaking, I want to know whether one-relator groups only have 'obvious' free splittings.

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 $F$?

Remarks


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*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.)

*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.)

*There is no decision-theoretic obstruction.  Magnus famously solved the word problem for one-relator groups. Much more recently, Nicholas Touikan has shown 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.

*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.

 A: Your question is reminiscent of Jaco's lemma. 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. 
A: 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 $x_1,\ldots,x_n$.
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?
A: $\newcommand{\rank}{\operatorname{rank}}$I think Grushko plus the Freiheitssatz does the trick.
Suppose that $G=A \ast B$ is a one-relator group which splits as a free product non-trivially. By Grushko, $\rank(G)=\rank(A)+\rank(B)=m+n$, where $\rank(A)=m$ and $\rank(B)=n$. If $G$ is not free, then by Grushko there is a one-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 one-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 $R$ 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.
A: I just came across the following, which is Prop. II.5.13 of Lyndon-Schupp.
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 $x_{k+1}, \ldots ,x_n$.
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.
