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For what possible values of n is U(n) cyclic?Is there any general result regarding this?If yes please could anyone state it and possibly give me a hint about its proof.

J. W. Tanner
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Shash
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  • What is your definition of $U(n)$? Usually it denotes the unitary group, but I suppose this is not the case here. Please add your own thoughts/efforts on the problem so that we could have some clue of how to help you. – WhatsUp Aug 24 '20 at 23:58
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    By $U(n)$, do you mean $(\mathbb Z/n\mathbb Z)^\times$? – J. W. Tanner Aug 25 '20 at 00:19
  • For $m\ge3$, $(\mathbb Z/2^m\mathbb Z)^\times$ is not cyclic, because $\pm1$ and $2^{m-1}\pm1$ are four distinct elements of order $2$ – J. W. Tanner Aug 25 '20 at 00:26
  • @J.W.Tanner slight correction. $(\Bbb Z/8\Bbb Z)^\times\cong \Bbb Z_2\times\Bbb Z_2$, so only has three elements of order $2$. –  Aug 25 '20 at 00:31
  • You are correct, @ChrisCuster; I suppose I meant of order $1$ or $2$ – J. W. Tanner Aug 25 '20 at 00:42
  • As we know, a cyclic group has at most one element of order $2$. –  Aug 25 '20 at 01:48
  • U(n) is the set which consists of all natural numbers which are less than n and relatively prime to n.Thus the group is U(n) along with multiplication Modulo n – Shash Aug 25 '20 at 02:45
  • Sorry that I didn't mention it clearly.I thought it was a well known notation just like Zn. – Shash Aug 25 '20 at 02:46

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The group is cyclic if and only if $n=1,2,4,p^k$ or $2p^k$ where $p$ is an odd prime.

Gauss knew this. It seems to me the result is fairly nontrivial, requiring the proof that there is a primitive element mod $n$ in those, and only those, cases. If memory serves this problem was at least partially considered in his Disquisitiones Arithmeticae of $1801$.