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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.

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    $\begingroup$ 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? $\endgroup$
    – Ian Agol
    Commented May 8, 2014 at 3:45
  • $\begingroup$ More infinitely generated examples similar to $Homeo(\mathbb R)$ can be found in scirp.org/journal/PaperDownload.aspx?paperID=3283 $\endgroup$
    – Igor Belegradek
    Commented May 8, 2014 at 11:54

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One is naturally led to consider the simple case, and I wonder if Graham Higman's 1951 examples might suffice. In J. Lond. Math. Soc. 26 (1951), 61-64, Higman gives examples of groups presented using 4 generators with 4 relations. (One can replace 4 by a larger integer.) According to these relations, each of the four generators is conjugate to its square. Of course, this is no guarantee that every group element has this property. Maybe someone can add why Higman's examples do or do not answer the question.

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    $\begingroup$ 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 $L=0$. $\endgroup$
    – YCor
    Commented May 9, 2014 at 19:07
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented May 9, 2014 at 19:09

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