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My question is pretty much as stated in the title: what examples are there (or does there exist) uncountable subgroups of a group but which have countably-infinite many cosets. I can only think of examples in the other direction (countable subgroups giving uncountably-many cosets).

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  • $\begingroup$ Your question seems to be ill-posed: for cosets you need a subgroup, so perhaps you want an uncountable group with an uncountable subgroup of infinite-countable index? $\endgroup$ – DonAntonio Sep 27 '13 at 4:43
  • $\begingroup$ @DonAntonio Isn't that what OP explicitly asked for? $\endgroup$ – anon Sep 27 '13 at 4:49
  • $\begingroup$ I don't think so, @anon, unless you can make sense of "[groups]...giving infinitely-countable many cosets" . As far as I kno,w, a group "doesn't give cosets", and there's always a subgroup involved. $\endgroup$ – DonAntonio Sep 27 '13 at 4:50
  • $\begingroup$ OP says "uncountable subgroups which give [i.e. yield or have] [countably-infinitely] many cosets." $\endgroup$ – anon Sep 27 '13 at 4:51
  • $\begingroup$ The question is all the words in it, not only part of them. $\endgroup$ – DonAntonio Sep 27 '13 at 5:09
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The uncountable subgroup of $\Bbb Z^{[0,1]}$ of maps $[0,1]\to\Bbb Z$ vanishing at $0$ has index $|\Bbb Z|=\aleph_0$.

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How about $\mathbb{R}$ in $\mathbb{Q} \times \mathbb{R}$? Or take any countably infinite group $H$, uncountable group $K$ and consider $K$ in $H \times K$.

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You can take the group $SO(n)$ inside $O(n)$. It has only two cosets!

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