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$G$ is a group with subgroups $H$ and $K$ such that $,H \cong K$, then is $G/H \cong G/ K$?

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Sona, don't you want $H$ and $K$ to be normal subgroups? What is the source of the problem? – Jonas Meyer May 31 '12 at 8:52
A related question on MathOverflow asks the converse question:… – Jonas Meyer May 31 '12 at 9:29
I wrote a long answer here on a specific case of the converse question a wee while back (Hopficity). – user1729 Jun 1 '12 at 11:14
This question is, essentially, a duplicate of this one, I believe (which floated to the top today). I like the $\mathbb{Z}$ example though. Perhaps the questions could be merged? – user1729 Jun 1 '12 at 11:23
Answered by/ Possible Duplicate of… – Eric Naslund Jun 1 '12 at 17:26
up vote 18 down vote accepted

No. Consider $G = (\mathbb{Z},+)$, $H= (2\mathbb{Z},+)$ and $K= (4\mathbb{Z},+)$. Note that $H$ and $K$ are isomorphic by the mapping $z \to 2z$.

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Wow, that was fast! (1 minute, 14 seconds after posting, complete with a link to a related question, although that might have come within the 5 minute window.) – Jonas Meyer May 31 '12 at 8:42
@JonasMeyer: Because first I typed, No. Consider $G=Z$, $H=2Z$, $K=4Z$. Then added more details. Does that take more than a minute :) – user9413 May 31 '12 at 9:23
Oh, I suppose that makes it less impressive :) – Jonas Meyer May 31 '12 at 9:25
@JonasMeyer: I am not here to impress anyone here with my typing speed :) – user9413 May 31 '12 at 9:26

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