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Why is every finite subgroup of $\mathbb C$* is cyclic?

Suppose on the contrary this it is false, then there exists a finite subgroup $G\subset \mathbb C$* with $o(G)=n$ such that $G$ is not cyclic. If $g$ is any non identity element, then $g^k=1$ for some $k|n$. Suppose that $k<n$. Then there is $g'$ in $G$ not in $(g)$, the cyclic group generated by $g$.

But I am not sure how to go from here to get a contradiction.

Koro
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    There is nothing special about $\mathbb{C}$. For any field $F$, a finite subgroup of $F^{\times}$ is cyclic. There are many ways to prove this. See here: https://math.stackexchange.com/questions/59903/finite-subgroups-of-the-multiplicative-group-of-a-field-are-cyclic – Mark Feb 27 '23 at 21:49
  • … but there probably is a simpler proof for the complex field – lhf Feb 27 '23 at 21:50
  • @Mark: can you please explain why in the linked answer cardinality of $\langle y\rangle $ is $d$? I think that we can only say that the cardinality is $\le d$. – Koro Feb 27 '23 at 21:57
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    @Koro Because $y\in G_d$, and $G_d$ was defined as the set of elements which have order $d$. (and the order of $y$ is equal to the cardinality of $\langle y\rangle$) – Mark Feb 27 '23 at 22:00
  • @Mark: Thanks a lot. I understand :-). – Koro Feb 27 '23 at 22:02
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    Here is the simpler proof I have in mind. By Lagrange’s theorem, $G$ is contained in the set of $n$th roots of unity, which is a cyclic group. Therefore, $G$ is cyclic. – lhf Feb 27 '23 at 23:38

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