This is a question I made up, but couldn't solve even after some days' thought. Also if any terminology is unclear or nonstandard, please complain.

Given groups $G$ and $H$, we say that $G$ can be embedded in $H$ if there exists an injective homomorphism $\varphi : G \to H$. (Note that the image $\varphi(G)$ is then isomorphic to $G$.) I am interested in the situation where a pair of groups $G$ and $H$ can be embedded in each other. Of course, this is guaranteed to be the case when $G \cong H$. But is the converse true? More precisely:

Q1. Do there exist non-isomorphic groups $G$ and $H$ such that each of them can be embedded in the other?

I am interested in this because, in my mind, this question is analogous to the Cantor-Bernstein-Schroeder theorem in set theory. Of course, this view could be too naive or useless. Oh well.

The only "progress" I could make is to create another question. Let $\varphi_G:G \to H$ and $\varphi_H:H \to G$ be a pair of embeddings as in the question. Then the homomorphism $\varphi := \varphi_H \circ \varphi_G : G \to G$ is also injective; i.e., it is an embedding. I can show that the image of this map ($K := \varphi(G)$) is a proper subgroup of $G$ unless $G \cong H$. This leads me to another question:

Q2. Does there exists a group $G$ that is isomorphic to a proper subgroup of itself?

If the answer to this is negative, then so is the case for Q1. Though both of these seem "obviously false", I cannot prove them. Nor can I construct a counterexample. Any suggestions?

Some remarks:

  • Nothing is inherently special about groups here. I suppose one could ask the same question for rings, fields, or other structures; I focused on this specific question for clarity.

  • I tried to search through Wikipedia and Google books, but I cannot figure out the answer or where I can find the answer.

  • I have no idea as to how easy or difficult these questions are. If they are trivial/easy (say, the level of a standard undergrad homework exercise), then please give me hints rather than a complete solution :-).

  • $\begingroup$ Look at non-abelian free groups of different ranks. Also, Hopfian and co-Hopfian groups seem of interest here. $\endgroup$ – t.b. Sep 7 '11 at 21:09
  • $\begingroup$ See here on the Secret Blogging Seminar, and also this MO question. $\endgroup$ – Zev Chonoles Sep 7 '11 at 21:11
  • $\begingroup$ So I am asking for an example of a non-Hopfian group? $\endgroup$ – Srivatsan Sep 7 '11 at 21:12
  • 2
    $\begingroup$ Well, the answer to your second question is yes: $2\mathbb{Z}\cong \mathbb{Z}$. Off the top of my head, I'm not sure about your first question. $\endgroup$ – user5137 Sep 7 '11 at 21:16
  • 2
    $\begingroup$ Related to this (at least tangentially): this MO thread asking about categories in which "Cantor-Bernstein" would hold. $\endgroup$ – Arturo Magidin Sep 7 '11 at 21:23

Let $F$ be a free group of finite rank $r > 1$. Then the commutator subgroup $[F,F]$ of $F$ is a free group of (countably!) infinite rank. Similarly but more easily, a free group of countably infinite rank contains as subgroups free groups of all finite ranks.

From this it follows that for any $r_1, r_2$ with $2 \leq r_1, r_2 \leq \aleph_0$, $r_1 \neq r_2$, the free group of rank $r_1$ and the free group of rank $r_2$ can be embedded in each other.

Comment: It is a lot easier to find examples of groups which are isomorphic to proper subgroups of themselves (or, in fancier terminology, non co-Hopfian groups). For instance an infinite cyclic group has this property, as does any nontrivial free abelian group or any infinite-dimensional vector space over $\mathbb{F}_p$ or $\mathbb{Q}$. (Added after seeing Arturo's answer: or, more generally, an infinite direct sum of copies of any nontrivial group!)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.