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ADL
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Hyperbolic groups with infinitely generated commutator subgroups
It is an old result (from "On one-relator groups having elements of finite order" by J. Fischer, A. Karrass and D. Solitar (1972)), but if $G=\langle X; R^n\rangle$, $n>1$, then the derived subgroup is an infinite free product of cyclic groups of order $n$. Also, $G$ is hyperbolic (and has one end).
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Cubical Complexes and Bass-Serre theory
Hmm - I really didn't pick up on this! (But it certainly makes sense - thanks.) Sageev talks about "essential" actions (which corresponds to not messing up the hyperplanes, and in the finite dimensional case this corresponds to an action with an unbounded orbit) - was this dropped somewhere along the way?
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When is an HNN-extension finitely presented?
Clarifying things a bit, as discussed in the comments.
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When is an HNN-extension finitely presented?
@HW: The sort of answer I would hope for would be "If you put these restrictions on $G$ and $K$ (and perhaps on the isomorphism between $K$ and $K^{\prime}$) then you can say something." Along the lines of the Baumslag-Tretfoff conditions for residual finiteness; sufficient, but not necessary.
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Any method to detect subgroup generated by a subset of the generators from its presentation
In general, you can use the reidemeister-schreier algorithm to compute the presentation of a subgroup of a given subgroup $H\leq G$. It works by looking at the cosets $G/H$, so it is nicest if $H\lhd G$. If $H$ has finite index in $G$ then it will spit out a finite presentation for $H$. I found a worked example of this here, math.stackexchange.com/questions/59273/…
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When are isomorphic copies of the base group in an $HNN$-extension subgroups of the base group (up to conjugacy)?
So...groups with property FA are the only groups which have the property in my question? (I.e. having property FA is equivalent to the property I am wondering about?)
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