3

Suppose we have two groups, $G,H$ and that $\rho:G\rightarrow H, \tau:H\rightarrow G$ are $1-1$ morphisms. Is it true then that $G,H$ are isomorphic? I don't think so, but I can't come up with any counter-example. This definitely holds for finite groups obviously. But what about infinite groups? Moreover, if it doesn't hold in general, does it hold for some large class of groups?

Just dropped in
  • 16,664
  • 3
    A group is coHopfian if it is not isomorphic to a subgroup of itself. If either of $G$ or $H$ is then it is true that $G\cong H$. One-ended hyperbolic groups are coHopfian, and indeed "almost all" groups are coHopfian (where "almost all" has a specific meaning, defined by Gromov). – user1729 Aug 27 '19 at 15:15

1 Answers1

8

One counterexample: the free group on two letters contains a subgroup isomorphic to the free group on three letters. (see here: Show that the free group on three generators is a subgroup of the free group on two generators)

hunter
  • 32,629