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
- 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.
Best Answer
$\newcommand{\rank}{\operatorname{rank}}$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.