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I know it is possible for a group $G$ to have normal subgroups $H, K$, such that $H\cong K$ but $G/H\not\cong G/K$, but I couldn't think of any examples with $G$ finite. What is an illustrative example?

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Take $G = \mathbb{Z}_4 \times \mathbb{Z}_2$, $H$ generated by $(0,1)$, $K$ generated by $(2,0)$. Then $H \cong K \cong \mathbb{Z}_2$ but $G/H \cong \mathbb{Z}_4$ while $G/K \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.

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  • $\begingroup$ Ah, thanks, that's a good one. Clearly my group theory is a little rusty... $\endgroup$ – Zev Chonoles Oct 24 '10 at 20:38
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    $\begingroup$ Nice example! $\endgroup$ – anonymous Oct 25 '10 at 6:16
  • $\begingroup$ @Nate: This will also work if our $H = \mathbb{Z}_{2}$ right? $\endgroup$ – user9413 Oct 14 '11 at 19:47
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    $\begingroup$ @Chandrasekhar: I'm not sure what you're asking. You can think of $H$ as $\{0\} \times \mathbb{Z}_2 \subset \mathbb{Z}_4 \times \mathbb{Z}_2$ if you want... $\endgroup$ – Nate Eldredge Oct 14 '11 at 21:30
  • $\begingroup$ @NateEldredge: thats precisely what i wanted. $\endgroup$ – user9413 Oct 14 '11 at 21:53

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