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Give an example of two groups $G_1$, $ G_2$ such that $G_1$ embed in $G_2 $ and $G_2$ embed in $G_1$ ($H$ embed in $G$ means that there exist $K$ a subgroup of $G$ s.t. $K$ and $H$ are isomorphic), but $$G_1≇G_2.$$ ($G_1≇G_2$ $⟺$ $\nexists f:G_1\to G_2$, $f$ 1-1, onto and homomorphism.)

Thanks in advance.

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I'm interested to know why you accepted the answer about free groups, when apparently you haven't learnt about free groups? It's a good answer, but surely not the answer expected if this is a homework problem for a course which hasn't covered free groups yet. –  Tara B Feb 26 '13 at 17:33
    
So how do you know the answer is true? –  Tara B Feb 26 '13 at 17:46
    
Sure, but all that doesn't really help with your homework or your understanding, does it? –  Tara B Feb 26 '13 at 17:58
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it doesn't help me now ,but it force me to study about free group. and im sure i will understand it –  Maisam Hedyelloo Feb 26 '13 at 18:06
    
Ah, okay, great, if you're going to follow it up and figure out why it's true for yourself, then that's fine! (And in fact the answer does help, because it tells you what to learn about.) –  Tara B Feb 26 '13 at 18:13
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4 Answers

up vote 7 down vote accepted

Take two free groups of rank 2 and 3.

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+1 Nice hint $\star$ –  B. S. Feb 26 '13 at 16:58
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How about something like this:

$$G_1=\mathbb Z_5 \times \mathbb Z_{5^2}\times \mathbb Z_{5^3}\times \mathbb Z_{5^4} \times \cdots$$ $$G_2= \mathbb Z_{5^2}\times \mathbb Z_{5^3}\times \mathbb Z_{5^4} \times ...$$

$G_2$ is embedded in $G_1$, since $G_2 \cong \{0\} \times \mathbb Z_{5^2}\times \mathbb Z_{5^3}\times \mathbb Z_{5^4} \times \cdots$ ; and $G_1$ is embedded in $G_2$ since $G_1 \cong 5G_2$.

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I like this very much, because I'm fairly sure it doesn't require anything beyond what the OP would have already learnt. –  Tara B Mar 2 '13 at 20:59
    
@TaraB This kind of examples are classical: see, for instance, Kaplansky's book Infinite Abelian Groups, page 13. –  user26857 Mar 2 '13 at 21:09
    
@YACP: Yes, sure, it's an obvious idea. I just wanted to comment because I thought it was a particularly suitable one, and other than commenting I couldn't do more to express my approval than upvote. –  Tara B Mar 2 '13 at 21:11
    
@YACP, thanks for the reference. –  Ludolila Mar 2 '13 at 21:13
    
@YACP: Wasn't 'Welcome' addressed to Ludolila for the thanks? –  Tara B Mar 2 '13 at 21:15
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By using the theorem of classification of divisible groups, we can easily proved that the claim is true, but if we omit the adjective divisible, the claim fails. If fact, as @Boris noted correctly, by taking two free groups of different rank$>2$ or equal to $2$ you can see the claim proved. See this also Is this statement true for divisible Groups? and Exercises in Abelian Group Theory page 155 for more.

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@MaisamHedyelloo: Hi, In what source did you face this problem? Some problems might have answers in the other area where you had expected. –  B. S. Feb 26 '13 at 17:09
    
@MaisamHedyelloo: Note that we will have troubles if we want to find such these groups among finite groups. Right? So, we look for an infinite structures. –  B. S. Feb 26 '13 at 17:36
    
Nice observation and clarification! –  amWhy Feb 26 '13 at 17:48
    
@BabakS.: The way you've writted this answer isn't that clear, because there isn't really a 'claim' in the question, but insofar as there is one, it's that there do exist two non-isomorphic groups which are isomorphic to subgroups of each other. And you are using 'the claim' to mean the opposite. –  Tara B Feb 26 '13 at 18:18
    
@TaraB: I found this question in the book of Rotman. There, he wanted me to prove the claim just for divisible groups and in the rest; he asked me to verify what would be happened if we omited the divisibility. Thanks –  B. S. Feb 28 '13 at 20:09
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A simple example is $\mathbb{Z}$ with addition and the even numbers $\{2 x \colon x \in \mathbb{Z}\}$. The second is a proper subgroup of the first, but isomorphic to it anyway. Clearly any example has to be infinite.

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The question asks for non-isomorphic groups. –  Jim Feb 26 '13 at 17:53
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