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It's mentioned in herstein that there can be infinite groups whose subgroups have finite index. I cannot think of any examples. Some examples would be useful

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  • $\begingroup$ A related interesting phenomenon is that infinite groups can have subgroups of finite order. The group of non-zero real numbers under multiplication incorporates the subgroup $\{-1, 1\}$ which I thought was quite amazing. $\endgroup$
    – Ishfaaq
    Dec 25, 2015 at 10:56
  • $\begingroup$ I am surprised that you cannot think of any examples, since the infinite cyclic group, which must be just aboout the best known infinite group, is an example. $\endgroup$
    – Derek Holt
    Dec 25, 2015 at 11:08
  • $\begingroup$ Yes. I was making a foolish mistake $\endgroup$
    – Non-Being
    Dec 25, 2015 at 11:51
  • $\begingroup$ The trivial subgroup has infinite index in any infinite group... $\endgroup$
    – YCor
    Dec 28, 2015 at 0:45

1 Answer 1

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The even integers in the integers. Any nontrivial subgroup of the integers.

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  • $\begingroup$ How is it finite? $\endgroup$
    – Non-Being
    Dec 25, 2015 at 10:57
  • $\begingroup$ It has finite index, which is what you asked about. $\endgroup$
    – verret
    Dec 25, 2015 at 11:06
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    $\begingroup$ Formally the index of a subgroup is equal to the number of cosets. $\mathbb{Z}/\mathbb{2Z}$ will have two cosets, namely: $0+\mathbb{2Z}$ and $1+\mathbb{2Z}$, therefore the index is 2. $\endgroup$
    – JukesOnYou
    Dec 25, 2015 at 11:08
  • $\begingroup$ Oh yess! I was confusing it with infinite order of each coset. A very foolish mistake. $\endgroup$
    – Non-Being
    Dec 25, 2015 at 11:50

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