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Is it possible to have a group $G,$ which has two different, but isomorphic subgroups $H$ and $H',$ such that one is of finite index, and the other one is of infinite index? If not, why is that not possible. If there is a counterexample please give one.

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Nonabelian free groups give finitely generated examples, as opposed to taking infinite products as suggested in the answers. –  t.b. Sep 10 '12 at 9:14
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The other examples are infinitely presented (indeed, generated). That is, they are of the form $G=\langle X; R\rangle$ where $X$ or $R$ is infinite. In group theory, the challenge is often to find finitely presented groups with extreme properties because it is "easy" to find infinitely presented groups (indeed, finitely-generated infinitely-presented groups!) with extreme properties$^{\ast}$. For example, there exist infinite finitely generated groups where every element has finite order, but all known examples are infinitely presented...or hopfian groups - it is easy to reel of a non-Hopfian group, but finding one which is finitely presented is harder (indeed, the other two answers both give you non-hopfian groups). See this question for the definition and examples of non-Hopfian groups.

So, I now need to give you a finitely presented example. A few things we need to know first, and they are all well-known and proofs should all be easily obtainable (they should all be in Magnus, Karrass and Solitar's book "Combinatorial Group Theory", or can be found on this here website, or you can try and prove them yourself - one of them is pretty easy...).

  • If a group is finitely presented then every subgroup of finite index is finitely presented,
  • Subgroups of free groups are free,
  • If $G$ is a free group of rank at least two then $G^{\prime}$ is the free group of countable rank, and is of infinite index.

Okay, let $G=F_n$ be the free group on $n$ generators, $n\geq 2$. Then $G$ has subgroups of finite index (why?) and these are all finitely presented and free. So, take one of them and call it $F_m$ (as it is free, on $m$-generators). On the other hand, take $m$-elements of the basis of $G^{\prime}$. These are both $m$-generated, and as subgroups of free groups are free and free groups are isomorphic if and only if they are free on the same cardinality of generators we are done!

$\ast$: I say "easy", but I really mean "easier". For example, Ol'shanskii proved that there exist infinite groups where every subgroup has order $p$ for some fixed, large enough prime $p$. His construction gives a two-generated, infinitely presented group. He has a book, which is pretty much 500 pages long, and is basically just the proof. It isn't light reading, either (it is called "Geometry of Defining Relations in Groups").

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And how do you know that your $\,F_m\,$ and that group generated by $\,m\,$ elements in the commutator subgroup one is of finite index and the other one isn't? –  DonAntonio Sep 10 '12 at 9:31
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@DonAntonio: $F_m$ is of finite index, by assumption. The other subgroup is of infinite index as it is the subgroup of a subgroup of infinite index. If $G<H<K$ then $|G:K|\geq |H:K|$ (with equality if and only if $H$ is of infinite index, because we took the subgroups to be proper as I am too lazy to type $\leq$). –  user1729 Sep 10 '12 at 9:44
    
For your first paragraph there's also a relevant MO thread Finitely presented infinite group with no element of infinite order? –  t.b. Sep 10 '12 at 9:50
    
@user1729 Thanx. I misread. Clear now +1 –  DonAntonio Sep 10 '12 at 10:17
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$\mathbb Z^{\mathbb N}$ has itself as subgroup of infinite index. That is: Let $G$ be the (abelian) group of all maps $f:\mathbb N\to \mathbb Z$. Let $H=G$ and $H'=\{f\in G\mid f(1)=0\}$. Then $G/H\cong 1$, $G/H'\cong\mathbb Z$ and $\phi:H'\to H$ given by $\phi(f)(x)=f(x+1)$ is "obviously" an isomorphism.

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Yes. For $n\in\Bbb N$ let $G_n$ be a copy of $\Bbb Z/2\Bbb Z$, and let $G=\prod_{n\in\Bbb N}G_n$ be the direct product. Then $H_0=\{0\}\times\prod_{n>0}G_n$ is isomorphic to $H_1=\prod_{n\in\Bbb N}A_n$, where $A_n=\{0\}$ if $n$ is odd and $A_n=G_n$ if $n$ is even. Clearly $[G:H_0]=2$ and $[G:H_1]$ is infinite.

Of course $H_0$ and $H_1$ are both isomorphic to $G$, so I could have used $G$ instead of $H_0$, but I thought that you might like the subgroups to be proper.

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