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$ U_ {p} $ is cyclic for p is a odd prime.

I saw this question. Help to prove that $ U_{p} $ is a cyclic group. And I understood that the statement -- If $G$ be a group of order $n$ and there exists at most one subgroup of order of every divisor of $n$ , then $G$ is cyclic.

But what to do after that to prove $ U_ {p} $ is cyclic for p is a odd prime? Can anyone please help me?

cmi
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    It is cyclic mainly because $\Bbb{Z/pZ}$ is a field so $x^n-1$ has at most $n$ roots. In $\Bbb{Z/15Z}$ then $x^2-1$ has $4$ roots – reuns Mar 24 '19 at 15:19
  • Ohh then it is easy. Can we prove it in similar manner for the case of $U_{p^k}$?@reuns – cmi Mar 24 '19 at 15:23
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    @cmi $\Bbb{Z}/p^k\Bbb{Z}$ is not a field. – jgon Mar 24 '19 at 15:24
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    If you meant $(\Bbb{Z/p^kZ})^\times$ then for $p$ odd prime it is cyclic because $(1+ p)^{p^m} \equiv 1+ p^{m+1} \bmod p^{m+2}$ so $1+p$ is of order $p^{k-1}$. For $p=2$ it is $(1+ 2)^{p^m} \equiv 1+ 2^{m+1}+\frac{2^m (2^m-1)}{2} 2^2 \bmod 2^{m+2}$ – reuns Mar 24 '19 at 15:25
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    @cmi Keith Conrad has written a very nice article on $U_p$. Here is the link: https://kconrad.math.uconn.edu/blurbs/grouptheory/cyclicmodp.pdf Kindly go through. You will find everything about $U_p$ – C.S. Mar 24 '19 at 15:25
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    You will find almost everything also on this site. You can also start here. – Dietrich Burde Mar 25 '19 at 21:42
  • I understood it...@DietrichBurde – cmi Mar 27 '19 at 12:27

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