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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?2012-05-31
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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-isomorphic2012-05-31
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    I wrote a [long answer](http://math.stackexchange.com/questions/79852/does-g-cong-g-h-imply-that-h-is-trivial/79907#79907) here on a specific case of the converse question a wee while back (Hopficity).2012-06-01
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    This question is, essentially, a duplicate of [this one](http://math.stackexchange.com/questions/7720/finite-group-with-isomorphic-normal-subgroups-and-non-isomorphic-quotients), I believe (which floated to the top today). I like the $\mathbb{Z}$ example though. Perhaps the questions could be merged?2012-06-01
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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...)2012-06-01
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    Answered by/ Possible Duplicate of http://math.stackexchange.com/questions/7720/finite-group-with-isomorphic-normal-subgroups-and-non-isomorphic-quotients2012-06-01

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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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    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.)2012-05-31
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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 :)2012-05-31
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    Oh, I suppose that makes it less impressive :)2012-05-31
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    @JonasMeyer: I am not here to *impress* anyone here with my typing speed :)2012-05-31