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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 :-).

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Look at non-abelian free groups of different ranks. Also, Hopfian and co-Hopfian groups seem of interest here. –  t.b. Sep 7 '11 at 21:09
    
See here on the Secret Blogging Seminar, and also this MO question. –  Zev Chonoles Sep 7 '11 at 21:11
    
So I am asking for an example of a non-Hopfian group? –  Srivatsan Sep 7 '11 at 21:12
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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. –  user5137 Sep 7 '11 at 21:16
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Related to this (at least tangentially): this MO thread asking about categories in which "Cantor-Bernstein" would hold. –  Arturo Magidin Sep 7 '11 at 21:23
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1 Answer 1

up vote 6 down vote accepted

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!)

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