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Since my reputation is not enough to comment on the original question, I have to post a question here, I will sincerely apologize if this behavior is not appropriate.

In the comments and the answer, @Derek Holt said that $|G:H|=2$ follows from the fact the automorphism group of an infinite cyclic group has order 2 , and for $y \in G\setminus H$, $y^{-1}xy = x^{-1}$, but I still cannot understand it.

I know that if $H$ is normal in $G$ then there is a homomorphism from $G$ to $Aut(H)$, hence $|G:H| \leq 2$, but how to prove that $H$ is normal? Or there is a way to show this without using normality?

Quzs
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    A subgroup of index 2 is always normal. – Arturo Magidin Mar 28 '23 at 12:35
  • @ArturoMagidin I know this, but my central question is that I cannot understand how to know $H$ is of index $2$, and I think that he deduces this from the normality, is there anything wrong with my thoughts? – Quzs Mar 28 '23 at 12:42
  • I think you have might a point -- I added a comment on Derek's answer. But this question should be closed anyway, in my opinion. Also see the older answer at https://math.stackexchange.com/questions/379234/terminology-for-infinite-groups-all-of-whose-subgroup-have-finite-index, which seems to be bullet-proof. – Sean Eberhard Mar 28 '23 at 13:18
  • @SeanEberhard Thank you for your help! I have read the older answer you mentioned, it is really good. – Quzs Mar 28 '23 at 13:24
  • Sorry, I assumed normality of $H$ in my answer. I will correct it. – Derek Holt Mar 28 '23 at 13:40
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    So the issue was addressed by replacing $H$ with its core, and is now solved in the original post. I think this should be closed as a duplicate pointing there. Agreed? – Arturo Magidin Mar 28 '23 at 14:09

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