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Do there exist groups $G$ and $H$ for which $G$ is isomorphic to a subgroup of $H$ and $H$ is isomorphic to a subgroup of $G$, but in fact $G$ is not isomorphic to $H$?

I know that $G = F_2$ and $H = F_3$, where $F_n$ denotes the free group on $n$ generators, satisfies this, but I can't convince myself that this is right. So, are there any "easier" or more "intuitive" groups that have the properties I wrote out above?

The reason for this question is because I am trying to find convincing arguments--not formal proofs, I understand those--of how free groups can have such unintuitive properties. How can I get a handle on the fact that the rank of a free group be well-defined, while the rank of its subgroups can have ranks of any countable cardinal?

cxx
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    A mathematical, and in fact not so hard, proof won't convince you? Any example will be out of infinite groups, and among these one free groups are in some rather intuitive sense simpler... – DonAntonio Mar 26 '20 at 23:39
  • Just curious how do you get $F_3$ in $F_2$? – Gregory Grant Mar 26 '20 at 23:43
  • See https://math.stackexchange.com/questions/3123098/show-that-the-free-group-on-three-generators-is-a-subgroup-of-the-free-group-on – cxx Mar 26 '20 at 23:44
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    @GregoryGrant the words $x^2,xy, yx$ generate all even-length words in $F_2$, and by the Schreier-Nielsen Subgroup Thm., you get an isomorphic copy of $F_3$ in $F_2$. – janmarqz Mar 27 '20 at 00:30
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    As noted in above, any example of this phenomenon would have to be of infinite groups. This is down to the cardinality definition of an infinite set; viz., that a set $X$ is infinite iff there exists a strict subset $Y\subset X$ such that $|Y|=|X|$. Just note that isomorphisms preserve order. – Shaun Mar 27 '20 at 00:32
  • My hunch is to look at free products with amalgamation for more examples. I tried cooking one up but got nowhere. – Shaun Mar 27 '20 at 00:34
  • @cxx The rank of a subgroup of a free group is well defined because the subgroup theorem for free groups: If $H$ is a subgroup of a free group then $H$ is also free – janmarqz Mar 27 '20 at 00:58
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    @janmarqz There is some ambiguity in my wording. I mean that I find it hard to swallow that there are subgroups which have ranks of any countable cardinality, while the entire group itself has a well-defined rank. – cxx Mar 27 '20 at 01:09
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    counterintuitive? yes!... but that is why maths are alive and are they exciting! – janmarqz Mar 27 '20 at 04:52
  • How is your topology? One characterisation of free groups is that they are fundamental groups of graphs (indeed, if $\Gamma$ is a graph with maximal spanning tree $T$ then $\pi_1(\Gamma)$ is free of rank $n$, where $n$ is the number of edges in $\Gamma\setminus T$). Then covers correspond to subgroups, and it's not hard to cook up a rank three graph covering a rank two graph, etc. – user1729 Mar 27 '20 at 10:17
  • @user1729 Hmm. My topology isn't too great, but this sounds like something I could at least visualize. I guess I'll look into it. Thanks. – cxx Mar 27 '20 at 15:47

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