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May 9, 2014 at 19:09 comment added YCor 2) As long as the question concerns finitely presented groups, it is not true that one is led to the simple case (because any nontrivial fg group has a simple quotient, but for a finitely presented group, one does not necessarily get a finitely presented simple quotient); actually Miller's group (an infinite f.p. group all of whose nontrivial quotients have a nonsolvable word problem) has no finitely presented simple quotient at all.
May 9, 2014 at 19:07 comment added YCor 1) This Higman group doesn't work: it has an action on a tree with hyperbolic elements. Such a hyperbolic element $g$ cannot be conjugate to their square, because if $L$ is its displacement length then the displacement of $g^2$ is $2L$, whence $L=0$.
May 9, 2014 at 16:49 review First posts
May 9, 2014 at 16:49
May 9, 2014 at 16:32 history answered Eric Moorhouse CC BY-SA 3.0