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Apr 16, 2011 at 16:26 comment added JeremyKun No. It is definitely an HNN extension of a free group. I just posed the more general question because I didn't know of this theorem on HNN extensions, and thought to answer it from a different angle.
Apr 16, 2011 at 4:19 comment added Autumn Kent Note that my answer only works when $G$ is in fact an HNN-extension of a free group, which isn't obviously the case for all groups with the kind of presentation you've given. (For instance, the relations could be $t^{-1}x_i t = x_1$ for all $i$, and so $t$ isn't giving you an isomorphism between subgroups.) Could that be the source of the problem?
Apr 16, 2011 at 3:31 comment added JeremyKun This is unfortunate for at least one author who I'll leave anonymous, because it provides a counterexample to a theorem in one of his books on one relator groups.
Apr 15, 2011 at 4:23 vote accept JeremyKun
Apr 14, 2011 at 22:36 history answered Autumn Kent CC BY-SA 3.0