$G$ is a group with subgroups $H$ and $K$ such that $,H \cong K$, then is $G/H \cong G/ K$?
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1Sona, don't you want $H$ and $K$ to be normal subgroups? What is the source of the problem? – Jonas Meyer May 31 '12 at 08:52
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A related question on MathOverflow asks the converse question: http://mathoverflow.net/questions/17221/can-a-quotient-of-a-group-by-two-different-subgroups-be-isomorphic – Jonas Meyer May 31 '12 at 09:29
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I wrote a long answer here on a specific case of the converse question a wee while back (Hopficity). – user1729 Jun 01 '12 at 11:14
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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 01 '12 at 11:23
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(And I have just read the answer - it didn't have those links yesterday! This would explain the floating to the top-ness...) – user1729 Jun 01 '12 at 11:24
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2Answered by/ Possible Duplicate of http://math.stackexchange.com/questions/7720/finite-group-with-isomorphic-normal-subgroups-and-non-isomorphic-quotients – Eric Naslund Jun 01 '12 at 17:26
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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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2Wow, 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 08:42
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@JonasMeyer: Because first I typed, No. Consider $G=Z$, $H=2Z$, $K=4Z$. Then added more details. Does that take more than a minute :) – May 31 '12 at 09:23
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