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Suppose $N$ is a normal subgroup of a group $G$, with $N \neq \{e\}$. Can $G/N$ be isomorphic to $G$?

My attempt: It is not possible in the case when $G$ is finite; a quick way to see this is to note that $o(G/N) = o(G)/o(N)$, so $G/N$ and $G$ do not have the same number of elements. For the general case, I suspect that it isn't possible as well, but I'm not sure. My intuition is that since $G$ is isomorphic to $G/\{e\}$, it shouldn't be isomorphic to another $G/N$, but I'm not sure if this is accurate.

Martin Sleziak
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twosigma
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2 Answers2

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Let $G=\bigoplus_{j=1}^\infty \mathbb{Z}$, an infinite sum of infinite cyclic groups. Then, if $N$ is the (normal) subgroup consisting of the first summand, $G/N\simeq G$. As you have shown, it is not possible in the finite case.

Douglas Molin
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A group $G$ is called Hopfian if it is not isomorphic to any proper quotient $G/N$. The most well known finitely generated (even 1-related) non-Hopfian group was found by Baumslag and Solitar. It is the group given by two generators $a,b$ and one relator $ba^2b^{-1}=a^3$.

markvs
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    I don't think this was the first such example, although it is probably the easiest. According to Lyndon and Schupp, the first finitely generated (but infinitely related) example was due to B H Neumann in 1950, and the first finitely presented example (the 2-generator and 3-relator group $\langle a,b,c \mid c^a=c^b=c^2 \rangle$) due to G Higman in 1951. Higman believed that there could be no 1-relator example, and the Baumslag Solitar groups were constructed as a counterexamples. – Derek Holt Sep 11 '20 at 07:56
  • OK. Changed "first" to "the most well known". – markvs Sep 11 '20 at 10:54