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Let $A$ be a free group and $G = A*_t$. When is $G$ also a free group? Suppose $t y t^{-1} = z$ and there is a splitting $A = B*C$ so that $y \in B$ and $z \in C$ and $z$ is a member of some basis of $C$ then clearly $G$ is free. Is this the only case that $G$ is free?

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up vote 10 down vote accepted

Yes. This is a theorem of Shenitzer. For a modern treatment see, for instance, this recent paper of Louder. I give a proof of a similar fact in section 2 of this preprint.

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Thank you! I appreciate the help! I looked at Louder's paper and I don't quite understand the proof. Since the proof is given for the amalgamation - I'll revert to that case. One can construct immersions $\Gamma_n \rightarrow R_{n+m-1}$ and $\Gamma_m \rightarrow R_{n+m-1}$ and $t$ immerses in each graph. The composition of immersions is still an immersion so the loops representing $t$ under each map are immersions which are homotopic (since $F_{n+m-1} \cong F_n *_{t} F_m$). Thus the images are different parametrization of the same immersed loop. ---I'm running out of space -- to be continued – spring Aug 12 '11 at 14:16
Now looking at the space $X$ constructed by an annulus mapping into $\Gamma_n \cup \Gamma_m$ via the immersions of $t$, one can extend the immersion of the graphs to an immersion of $X$. Great. Now he says that if we look at preimages of midpoints of $R_{n+m-1}$ then we get a forest. But I don't see how it isn't just a collection of arcs. The boundary of the annulus is mapped to the same loop (possibly under different parametrization) so it is just a collection of arcs. Also, why this implies that one of the boundaries crosses some edge only once is unclear to me... many thanks for your time! – spring Aug 12 '11 at 14:23
ok. I feel stupid. Of course the edges get identified at the endpoints. Thanks for the references!! – spring Aug 12 '11 at 16:44
spring: there is a newer, easier to read version of 'recent paper...'. I'll send you a copy if you're interested. – nolte Aug 16 '11 at 3:46

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