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



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> 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 $F$?
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**Remarks**


1. 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.)
2. 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.)
3. There is no decision-theoretic obstruction. Magnus famously solved the word problem for one-relator groups. Much more recently, [Nicholas Touikan has shown](http://arxiv.org/abs/0906.3902) 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.
4. 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.



Free splittings of one-relator groups
$\newcommand{\rank}{\operatorname{rank}}$I think [Grushko](http://en.wikipedia.org/wiki/Grushko_theorem) plus the [Freiheitssatz](http://en.wikipedia.org/wiki/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.


