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I am reading Introduction to Functional Equations by Costas Efthimiou, and in the proof of Theorem 7.2 on page 119, he claims that "it is known in algebra" that no subgroup of the additive group $(\mathbb R, +)$ has index 2. I could not find a proof of this fact, so I would appreciate it if someone could provide me one.

cubicquartic
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    It's a divisible group. Any quotient group would have to be divisible hence it if it finite it is trivial. – lulu Aug 29 '22 at 20:14
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    any subgroup generated by one element is either $0$ or isomorphic to $\mathbb{Z}$ in fact it will be the group $r\mathbb{Z}$ with $r\neq 0$. But one has that for example $r\mathbb{Z}+r/3$ and $r\mathbb{Z}+2r/3$ are two cosets for example. In fact there are uncountably many cosets – MIO Aug 29 '22 at 20:15
  • @lulu Why are you leaving an answer in the comment section? – Arthur Aug 29 '22 at 20:15
  • @Arthur Searching for a duplicate. Haven't found one just for $\mathbb R$ yet. – lulu Aug 29 '22 at 20:15
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    That's understandable. But why did you leave an answer in the comment section? – Arthur Aug 29 '22 at 20:27

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