Equivalent conditions for coHopfian groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T10:01:18Z http://mathoverflow.net/feeds/question/65881 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65881/equivalent-conditions-for-cohopfian-groups Equivalent conditions for coHopfian groups jkun 2011-05-24T17:32:31Z 2012-10-28T17:05:33Z <p>A lot of work has been done on determining whether particular classes of groups are coHopfian, in particular the (sufficiently large) braid groups $B_n$ modulo center, and certain classes of 3-manifold groups.</p> <p>Are there any equivalent conditions for a group to be coHopfian? I'm aware this is a big open problem, so I'm specifically looking for established results or necessary conditions.</p> <p>Of course, sufficient conditions are helpful as well, but most of what I've read in terms of sufficient conditions is "$G$ can be realized as a member of a class X, which we prove is coHopfian," and this is not what I want.</p> http://mathoverflow.net/questions/65881/equivalent-conditions-for-cohopfian-groups/110884#110884 Answer by Mark Sapir for Equivalent conditions for coHopfian groups Mark Sapir 2012-10-28T06:50:25Z 2012-10-28T17:05:33Z <p>This is in fact related to <a href="http://mathoverflow.net/questions/51263/infinitely-many-solutions-of-a-diophantine-equation" rel="nofollow">this</a> question. Let $\phi$ be an injective but not surjective homomorphism $G\to G$. Then under very mild conditions (see below) the powers $\phi, \phi^2,...,\phi^n,...$ are not conjugate in $G$ for all $n$ (see the definition in the linked question). Indeed, suppose that there exists a $g$ such that $\phi^{i+j}(x)=g\phi^i(x)g^{-1}$ for every $x\in G$ for some $i,j>0$. Then applying $\phi$, we get $\phi^{i+j+1}(x) = \phi(g)\phi^{i+1}(x)\phi(g)^{-1} = g\phi^{i+1}(x)g^{-1}$. Therefore $a=g^{-1}\phi(g)$ commutes with $\phi^{i+1}(x)$ for every $x$. Suppose that centralizers of subgroups of $G$ that are isomorphic to $G$ are trivial (it is the mild condition I mentioned before, that holds for non-elementary hyperbolic, relatively hyperbolic, etc. groups). Then $a=1$, hence $g=\phi(g)$. But then $\phi^{i+j}(x)=\phi^i(g)\phi^i(x)(\phi^i(g))^{-1}=\phi^i(gxg^{-1})$ for every $x$. Hence $\phi^j(x)=gxg^{-1}$ for every $x$ and $\phi^j$ is surjective, a contradiction. </p> <p>This implies (see the question above again) that $G$ acts non-trivially on an asymptotic cone of itself. Thus, for example, if $G$ is hyperbolic with property (T), it is co-Hopfian because its asymptotic cones are $\mathbb{R}$-trees and a group with property (T) cannot act on an $\mathbb{R}$-tree non-trivially. Much stronger results can be found in Sela, Z. Endomorphisms of hyperbolic groups. I. The Hopf property. Topology 38 (1999), no. 2, 301–321 and Druţu, Cornelia, Sapir, Mark V. Groups acting on tree-graded spaces and splittings of relatively hyperbolic groups. Adv. Math. 217 (2008), no. 3, 1313–1367. </p>