Do two isomorphic normal subgroups yield the same quotient

Question

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 $G=\mathbb Z$?

Answer 1

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, $\mathbb{Z}_6\not\simeq S_3$.

Answer 2

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!).