If $G$ is a group and $N$ is a nontrivial normal subgroup, can $G/N \cong G$?

Question

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 $G$.

Answer 1

Take $G = \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus \ldots $ and $N = \mathbb{Z}\oplus \{0\}\oplus \{0\} \ldots$

Answer 2

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...)

Answer 3

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.

Answer 4

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.

Answer 5

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 $T_p/N\cong T_p$.

Answer 6

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.