While studying for my finals this question came across my mind. Is there a group $G$ and a proper subgroup $H$ such that $G/H\cong G$? I immediatly discarded a finite group or even the integers, as for a finite group the orders wouldn't match and for the integers, the proper subgroups are all of the form $n\mathbb{Z}$ with $n$ integer. Having in mind the first isomorphism theorem I'm looking for a surjective homomorphism with non-trivial kernel. That would give $G/\ker(f) \cong G$. Is there any example?
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1Hint: Take any group $G$, and consider the countably infinite direct product: $$G^\omega=G\times G\times G\times G\times\cdots$$. – Christian E. Ramirez Jan 11 '23 at 10:45
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Let $G$ be the additive group of the complex numbers and $H$ the real numbers. Then $G/H\cong H\cong G$. – user14111 Jan 11 '23 at 10:50
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1For an easy countable example, if $F$ is a free group on $\aleph_0$ generators, then there is a surjective homomorphism $h:F\to F$ which is not an automorphism. – user14111 Jan 11 '23 at 10:55
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1You're looking for non-Hopfian groups. – calc ll Jan 11 '23 at 13:32
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Take your favourite group $\tilde{G}$ and let $G = \prod_{n=1}^{\infty} \tilde{G}$. Let $f : G \to G$ be the shift $f(x_1, x_2, \ldots) = (x_2, x_3, \ldots)$. Then $f$ is a surjective group homomorphism with $\ker(f) \cong \tilde{G}$.
jl00
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