# Finitely presented group in which every element is conjugate to its square

Does there exist a nontrivial finitely presented group in which every element is conjugate to its square? Is this an open problem?

Motivation: Jahren proved in [Geom Dedicata (2010)] that if $M$ is a closed manifold of dimension $\ge 5$ such that $\pi_1(M)$ has an element not conjugate to its square, then the smooth pseudoisotopy space for $M$ is not connected.

Finitely generated examples exist: Osin constructed a f.g. infinite torsion free group in which all nontrivial elements are conjugate.

• A couple of remarks: $Homeo^+(R)$ has this property, so one could look for a finitely-presented subgroup (something like Thompson's groups) which might also have this property? Also, such a group is perfect (in fact, every element is a commutator) and has no finite quotients. Is there a well-known finitely-presented infinite group in which every element is a commutator? – Ian Agol May 8 '14 at 3:45
• More infinitely generated examples similar to $Homeo(\mathbb R)$ can be found in scirp.org/journal/PaperDownload.aspx?paperID=3283 – Igor Belegradek May 8 '14 at 11:54

• 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 $L=0$. – YCor May 9 '14 at 19:07