I have just learnt a theorem that says:

For the group of invertible elements $F^*$ of a field $F$, its finite subgroup is cyclic.

And an example is that for $F=\mathbb{C}$, we have the group generated by n-th roots of unity finite subgroups, and hence cyclic.

So I am now considering if the only finite subgroup of $\Bbb C^*$ are group generated by the n-th root of unity. If so, how may I prove that there is no other finite subgroup? If not, may I please ask for some other example? Thanks!

  • $\begingroup$ I don't really understand what you are asking for, or the theorem that you learnt. Is it that the multiplicative group of every FINITE field is cyclic? or every FINITE subgroup of $\mathbb{C}^*$ is cyclic? $\endgroup$ – Darío G May 15 '17 at 10:33
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    $\begingroup$ The correct claim is: any finite subgroup of the multiplicative group of any field is cyclic. $\endgroup$ – DonAntonio May 15 '17 at 10:33
  • $\begingroup$ @Wore I am asking to determine the finite subgroup of $\Bbb C$. $\endgroup$ – PropositionX May 15 '17 at 10:36
  • $\begingroup$ @PropositionX You surely meant that you're asking to determine the finite subgroups of $\;\Bbb C^*\;$ ...it has infinitely many such subgroups, whereas the additive group $\;\Bbb C\;$ has none but the trivial one. $\endgroup$ – DonAntonio May 15 '17 at 10:38
  • $\begingroup$ Isn't it obvious that a cyclic subgroup of order $n$ in ${\mathbb C}^*$ consists of $n$-th roots of unity? $\endgroup$ – Derek Holt May 15 '17 at 10:41


If $\;C\;$ is a finite subgroup of $\;\Bbb C^*\;$ , then there exists

$$n\in\Bbb N\;\;\;s.t.\;\;\;\text{for all}\;\;\;c\in C\;,\;\;c^n=1$$

  • $\begingroup$ So is that correct that: Every element in a finite subgroup should always have finite multiplicative order, so the elements in the subgroup can only be roots of unity. And as if $n_1\ne n_2$, the product of 2 roots of unity does not give another root of unity, so we can only have subgroups generated by a single nth roots of unity? $\endgroup$ – PropositionX May 15 '17 at 10:47
  • $\begingroup$ @PropositionX Yes to your first question (obviously), and also its conclusion. Why do you think the product of two roots of unity cannot be a root of unity again? In fact, they must be, otherwise they cannot belong in a multiplicative group ...Even if you meant "two roots of unity of different order", their product is again a root of unity of order the roots' orders' common multiple. $\endgroup$ – DonAntonio May 15 '17 at 11:39
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    $\begingroup$ Oh I see, I do mean different order and forgot the common multiple case. It is clear now. Thanks for patience! $\endgroup$ – PropositionX May 15 '17 at 11:48

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