MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What can be said about pairs of non-isomorphic groups which are epimorphic images of one another and which also embed into one another?

Can such pairs of groups be 'classified' in some sufficiently weak, but still non-trivial sense, or are they just too common to hope for anything like this?

Obviously, groups forming such pairs can neither be Hopfian nor co-Hopfian.

An example of such pair of groups consists of $\rm C_\infty \times \rm F_2^\infty$ and $\rm F_2^\infty$, where $\rm C_\infty$ denotes the infinite cyclic group and $\rm F_2$ denotes the (nonabelian) free group of rank 2.

share|cite|improve this question
A slightly simpler example: $C_2 \times C_4^\infty$ and $C_4^\infty$. – S. Carnahan Jan 18 '13 at 14:53
Indeed. -- However which one is 'simpler' depends on your notion of 'simplicity': in order to define a free group, you need no relations, while to define a finite cyclic group, you need 1! – Stefan Kohl Jan 18 '13 at 15:00
@Berlusconi: I am not a group theorist, but my impression is that it is too optimistic to hope for a classification (without further assumptions and special conditions). – Martin Brandenburg Jan 22 '13 at 9:41
The examples given so far are of the form $(A \times B^\infty, B^\infty)$, where $A$ both embeds into $B$ and is an epimorphic image of $B$. -- Can anyone give an example where the groups do not admit a decomposition into infinitely many direct factors, or which are at least not of the form $(A \times B^\infty, B^\infty)$? – Stefan Kohl Jan 22 '13 at 16:00
I believe you can modify the example so that the direct products become semidirect products. – Khalid Bou-Rabee Jan 28 '13 at 5:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.