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I am wondering whether it is possible or not thata: $G \cong H$, $G_0 \cong H_0$ but $G/G_0 \not\cong H/H_0 $. In fact I found tahat as exercise asking to find such example but I can not find any. What is more important I can not understand how can it be possible. I thought isomorphisc groups are in fact thesame object with maby different names so I would be grateful also for some comment connected with that doubt.

J.E.M.S
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  • It can even be done with $G=H$, and it has been, on this website, previously. Have a look to see if you can find it. – Gerry Myerson Aug 19 '13 at 12:03
  • I think it might be better to mark as a duplicate of the question without the finiteness restriction, which changes the flavour of the answers a bit. I also hadn't seen this question before so didn't realize it was (or was likely to be) a duplicate until after answering. I'm inclined to leave my answer here because it has some general statements which don't seem to appear in answers to the duplicate questions. – mdp Aug 19 '13 at 12:17

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A hint to help you work out the counterexample - all (non-trivial) subgroups of $\mathbb{Z}$ are isomorphic to $\mathbb{Z}$.

Conceptually, the problem is that while $G_0$ and $H_0$ might be isomorphic, they needn't sit inside $G$ and $H$ in the same way (by which I mean that maybe no isomorphism between $G$ and $H$ restricts to an isomorphism between $G_0$ and $H_0$). In the case that $f\colon G\to H$ is an isomorphism, and $H_0=f(G_0)$, you do get $G/G_0\cong H/H_0$, but in general the two isomorphisms needn't be related in this way.

mdp
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