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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 :-).

user1729
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Srivatsan
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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 07 '11 at 21:09
  • See here on the Secret Blogging Seminar, and also this MO question. – Zev Chonoles Sep 07 '11 at 21:11
  • So I am asking for an example of a non-Hopfian group? – Srivatsan Sep 07 '11 at 21:12
  • Thompson's Group $F$ contains a copy of $F \times F$. Thompson's groups provide counterexamples to almost everything. As for a group isomorphic to a proper subgroup of itself, consider the simplest infinite group you can think of :) – MartianInvader Sep 07 '11 at 21:13
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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. –  Sep 07 '11 at 21:16
  • Theo, Zev, MartianInvader, Thanks for your comments. Need some time to mull over the nice variety of examples. So, that makes my "obviously false" guess "obviously wrong", yes? :-) – Srivatsan Sep 07 '11 at 21:17
  • $G\times G \times G \times \cdots $ is isomorphic to its proper subgroup ${ 1 } \times G \times G \times G \cdots $. Check out http://en.wikipedia.org/wiki/Hopfian_group for related concepts. – Matthew Towers Sep 07 '11 at 21:18
  • @Jack, Well :). Thanks. – Srivatsan Sep 07 '11 at 21:18
  • @Srivatsan: No, in Q2 you are asking for examples of non-co-Hopfian groups; non-Hopfian groups need not be isomorphic to proper subgroups of themselves. For example, the Prufer $p$-group is isomorphic to every quotient of itself except the trivial group, but is isomorphic to no proper subgroup of itself. (While the infinite cyclic group is isomorphic to every nontrivial subgroup of itself, but is isomorphic to no proper quotient of itself). – Arturo Magidin Sep 07 '11 at 21:21
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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 07 '11 at 21:23
  • @Arturo That's seems interesting, thanks! In retrospect it would've been nice if I had phrased it that way. Thought it might be too silly :) – Srivatsan Sep 07 '11 at 21:26

1 Answers1

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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!)

Pete L. Clark
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