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I know $G/N$ is isomorphic to a proper subgroup of $G$ in this case, so the gut instinct I had was 'no'. But there are examples of groups that are isomorphic to proper subgroups, such as the integers being isomorphic to the even integers, so that reasoning doesn't work. However in this case the even integers are not a quotient of the integers.

edit: I realize now that $G/N$ is not necessarily isomorphic to a proper subgroup of $G$, just a subgroup of $G$.

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  • $\begingroup$ The same question has been asked > 10 times on math.SE. Next time you should use the search function or google. $\endgroup$ – Martin Brandenburg Nov 17 '13 at 12:14
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    $\begingroup$ @MartinBrandenburg Using what search terms...? Just plug in "quotient isomorphic to group" and sift through 8 pages? $\endgroup$ – Jack M Nov 17 '13 at 14:40
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    $\begingroup$ I did try to search, but as Jack M said, if you pick the wrong terms it doesn't come up. I apologize for the duplicate. $\endgroup$ – NotAwake Nov 17 '13 at 18:23
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Take $G = \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus \ldots $ and $N = \mathbb{Z}\oplus \{0\}\oplus \{0\} \ldots$

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    $\begingroup$ I like this :) :) $\endgroup$ – user87543 Nov 17 '13 at 8:48
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A group with this property is called nonHopfian. It used to be an open question whether there were finitely generated examples, and some examples were eventually found by Baumslag and Solitar.

The simplest one is the group ${\rm BS}(2,3)$ defined by the presentation $\langle x,y \mid y^{-1}x^2y=x^3 \rangle$, where $N$ is the normal closure of the element $r=x^{-1} y^{-1} xyx^{-1} y^{-1} xyx^{-1}$.

The relator $r$ is equivalent to the relation $x=(x^{-1} y^{-1} xy)^2$, and adding this as an extra relation gives

$\langle x,y \mid y^{-1}x^2y=x^3, x=(x^{-1} y^{-1} xy)^2 \rangle \cong$ $\langle x,y,w \mid w=x^{-1} y^{-1} xy, y^{-1}x^2y=x^3, x=w^2 \rangle \cong$ $\langle y,w \mid y^{-1}w^2y=w^3, y^{-1}w^4y=w^6 \rangle \cong G.$

You also have to prove that $r$ is not equal to the identity in $G$, which requires a bit of the theory of HNN extensions.

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Yes. Let $G$ be the additive group of the complex numbers, and let $N$ be the subgroup consisting of the real numbers.

Edit in response to comment by @GA316:

$(\mathbb C,+)/\mathbb R$ is clearly isomorphic to $(\mathbb R, +)$, and it is well known (but this requires the Axiom of Choice) that $(\mathbb C,+)\cong(\mathbb R,+)$.

My answer is a special case of the answer posted simultaneously by Asaf Karagila, considering $\mathbb C$ as a vector space over the field of rational numbers.

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  • $\begingroup$ Why the downvote? $\endgroup$ – bof Nov 17 '13 at 19:17
  • $\begingroup$ can you explain your answer little more? I cant understand. why the quotient group Isomorphic to $\mathbb{C}$? $\endgroup$ – GA316 Nov 19 '13 at 18:20
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Take any infinite dimensional vector space $V$, and let $W$ be any subspace of $V$ whose direct complement has the same dimension as $V$. For example, any finite dimensional subspace.

Then $V/W$ must have the same dimension as $V$.

(And now note that vector spaces are Abelian groups, and that subspaces are subgroups...)

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  • $\begingroup$ Yes, this answer is a generalization of yours. I began writing it before you posted your answer, though, after deleting the original example which was the same as Prahlad's (who beat me by two seconds...) $\endgroup$ – Asaf Karagila Nov 17 '13 at 9:00
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Another example: Fix a prime $p$ and let $T_p=\{z\in\mathbb{C}|\exists n\geq0\hspace{5pt} z^{p^n}=1 \}$.
Consider the map $\varphi: T_p\to T_p$ defined by $\varphi(z)=z^p$. Then $\varphi$ is a homomorphism with non-trivial kernel, $N=\{z\in \mathbb{C}|z^p=1\}$ and $\operatorname{Im}(\varphi)=T_p$, so $T_p/N\cong T_p$.

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    $\begingroup$ This example also disproves the statement made by the OP that in this case the whole group is isomorphic to its proper subgroup: every proper subgroup in $T_p$ is finite, hence not isomorphic to $T_p$. $\endgroup$ – Dan Shved Nov 17 '13 at 14:18
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As mentionned by Derek Holt, the question is equivalent to find non-hopfian groups. In fact, it is easy to prove

Claim: Let $m,n \in \mathbb{Z} \backslash \{0, 1, -1\}$ be two coprime numbers. Then the Baumslag-Solitar group $BS(n,m)=\langle a,b \mid ab^ma^{-1}=b^n \rangle$ is non-hopfian.

Let $\varphi : BS(n,m) \to BS(n,m)$ be the morphism sending $a$ to $a$ and $b$ to $b^m$; $\varphi$ is well-defined since $\varphi(ab^ma^{-1}b^{-n})=1$.

  1. $\varphi$ is surjective. Indeed, $a,b^m \in \mathrm{Im}(\varphi)$ and $b^n=ab^ma^{-1} \in \mathrm{Im}(\varphi)$. Because $m$ and $n$ are coprime, there exist $p,q \in \mathbb{Z}$ such that $pm+qn=1$, hence $$b= b^{pm+qn}=(b^m)^p \cdot (b^n)^q \in \mathrm{Im}(\varphi).$$

  2. $\varphi$ is not injective. Indeed, $[aba^{-1},b] \neq 1$ (using Britton's lemma for example) but $\varphi([aba^{-1},b])=1$.

A necessary and sufficient condition is given here to know when $BS(n,m)$ is hopfian or not.

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  • $\begingroup$ I am not sure this counts as "easy to prove". If you are given the map and a word in the kernel then it is easy to prove, so long as you know something about HNN-extensions. Coming up with the map and word is less trivial. Which is, I suppose, one reason why there was a gap of 11 years between Higman's three-generator, two-relator group (the previous "best result") and this neater example of G.Baumslag and Solitar. $\endgroup$ – user1729 May 1 '14 at 8:33
  • $\begingroup$ I completely agree, it is not easy to find the group, the morphism and the word in order to find examples of non-hopfian groups. However, the group, the morphism and the word are quite simple, and surprisingly, the proof turns out to be elementary. $\endgroup$ – Seirios May 1 '14 at 9:05

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