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Is every finite subgroup of $C^*=$ set of all non-zero complex numbers cyclic?

I see that the set $A_n=\{z:z^n=1\}$ is a subgroup of $C^{*}$.

Any element of $A_n$ is a solution of $z^n=1$.Now the solutions of $z^n=1$ for any $n\in \Bbb N$ are $e^{\frac{{2ki\pi}}{{n}}};1\le k\le n$ and hence the subgroup $A_n$ is generated by $a=e^{\frac{{2ki\pi}}{{n}}}$ and hence cyclic.

But is it the case that any finite subgroup of $C^{*}$ is of the form $A_n$ for some $n$?

I am having problems to answer this question.If it is true then it answers the original question.

Please help.

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  • $\begingroup$ You should put emphasis on "But is it the case that any finite subgroup of $C^*$ is of the form $A_n$ for some $n$?" if this is what you really want to know. Otherwise, this is a duplicate. $\endgroup$ – Ennar Feb 23 '17 at 12:28
  • $\begingroup$ It would be a good thing if the group of all nonzero algebraic numbers were cyclic. $\endgroup$ – Marc Bogaerts Feb 23 '17 at 18:35
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All finite subgroups are cyclic. In fact this statement is true for $F^*$ for any field $F$. However one can easily construct infinite subgroups that are not cyclic. Consider all elements of finite order. This is a proper subgroup: it is precisely all the roots of unity. Because there are infinitely many prime numbers, one can see that not all numbers of the form $e^{2\pi i/p},\ p$ a prime, can be obtained as a power of a single complex number.

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  • $\begingroup$ Sorry for the wrong question;Its edited $\endgroup$ – Learnmore Feb 23 '17 at 11:08
  • $\begingroup$ For a proof see Serre's "A Course in Arithmetic", the first few pages. $\endgroup$ – P Vanchinathan Feb 23 '17 at 11:12
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$\mathbb C^\times$ is a subgroup of itself, and is certainly not cyclic.

If you're talking about proper subgroups only, consider the unit circle, or the group of nonzero complex numbers whose real and imaginary parts are both rational. Or $\mathbb R\setminus\{0\}$.

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  • $\begingroup$ Sorry for the wrong question;Its edited $\endgroup$ – Learnmore Feb 23 '17 at 11:08
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If $G$ is such finite subgroup, for every $g\in G, |g|=1$. Since if $|g|\neq 1$, $1,...,g^n,..., n\in N$ is infinite. You can write $g=e^{ic}$. Let $N$ be the order of $G$, $g^N=1$. This implies that $g^{Nic}=1$ and $C=p{2\pi\over N}$. We deduce that is a subgroup of the group generated by $e^{2i\pi\over N}$, and $G$ is cyclic since it is a subgroup of a finite cyclic group.

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  • $\begingroup$ This answer contains a mistake I think. How can you be sure that $H \leq \langle e^{2i\pi/N} \rangle$? $\endgroup$ – user370967 Oct 21 '17 at 14:18
  • $\begingroup$ You can fix the proof by writing $m:= lcm\{o(h) \mid h \in G\}$ and then $H \leq \langle e^{2\pi i/m}\rangle$ $\endgroup$ – user370967 Oct 21 '17 at 14:24
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    $\begingroup$ Also, the step $|g| = 1$ is necessary. You can immediately deduce this by $g^{ord(g)} = 1$ $\endgroup$ – user370967 Oct 21 '17 at 14:54

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