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Let $G$ be a group with normal subgroups $H\cong K$, then $G/H\cong G/K$ is not true in general. This question was being discussed here already. But I don't really understand why it's gotten so complicated there. Can't we just take $H=2\mathbb Z$ and $K=3\mathbb Z$ for $G=\mathbb Z$?

José Carlos Santos
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Buh
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    The question you're linking also gives a simple counter example...but it also goes way further into this matter. That is why, apparently, things get much messier there than simply giving a counter example. Read carefully the question there. – DonAntonio Feb 06 '18 at 08:56
  • Yes. This doesn't answer my question though. – Buh Feb 06 '18 at 08:58
  • Since the question over there was about finite groups, I don't see how your example answers it. – bof Feb 06 '18 at 09:04
  • @bof Now you're just trying to misunderstand me on purpose. Just read the first sentence again. – Buh Feb 06 '18 at 09:06
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    No, I must be misunderstanding you, but I'm not doing it on purpose. You linked to a question about finite groups, and you say you don't understand why it's so complicated there, and give an example with $G=\mathbb Z$. Call me dumbo, but I don't get it. – bof Feb 06 '18 at 09:17
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    Just because noone has said it yet: Yes, your counter-example does answer the question you ask. – user1729 Feb 06 '18 at 10:11
  • @user1729 Thank you. That's all I wanted to know. – Buh Feb 06 '18 at 10:35

2 Answers2

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Here's the simplest example that I was able to think of. Take $G=S_3\times\mathbb{Z}_3$. Let $H_1=\langle(1\ \ 2\ \ 3)\rangle\times\{0\}$ and let $H_2=\{e\}\times\mathbb{Z}_3$. Then $H_1\simeq H_2$, $G/H_1\simeq\mathbb{Z}_6$, and $G/H_2\simeq S_3$. Of course, $\mathbb{Z}_6\not\simeq S_3$.

José Carlos Santos
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  • You could also use two copies of $\mathbb{Z}/2\mathbb{Z}$ in $\mathbb{Z}/2\times \mathbb{Z}/4\mathbb{Z}$ (giving you respective quotients $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$). – user1729 Feb 06 '18 at 10:12
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Yes, your counter-example does answer the question you ask.

The question you link to asks a more technical question, involving automorphisms. That is why the discussion there is so much more complicated. Also, your counter-example does not answer the question there because it is assuming that the group $G$ is finite (and clearly $G\cong \mathbb{Z}$ is infinite!).

user1729
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