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Let $n$ be a given arbitrary positive integer, and let $U_n$ denote the group of all the positive integers less than $n$ and relatively prime to $n$ under multiplication mod $n$. Then for which values of $n$ is $U_n$ a cyclic group? And for any such $n$, which elements of $U_n$ generate it?

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    $\begingroup$ Exactly when $\;n=1\,,\,n=2\,,\,n=2p\,,\,p^k\;$ , with $\;p\;$ an odd prime. $\endgroup$ – DonAntonio Apr 2 '14 at 11:05
  • $\begingroup$ This would have been better if you include "I think" with "which, when" rather with only "which,when"... $\endgroup$ – user87543 Apr 2 '14 at 11:06
  • $\begingroup$ This is explained in almost all books on number theory. $\endgroup$ – lhf Apr 2 '14 at 11:06
  • $\begingroup$ Ihf, I'd appreciate if you could also give an exact reference where both parts of my question are answered. $\endgroup$ – Saaqib Mahmood Apr 2 '14 at 12:04
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    $\begingroup$ The problem of finding a generator for, say, $U_p$, when $p$ is a (large) prime, is not easy. There's no simple formula for it. $\endgroup$ – Gerry Myerson Apr 2 '14 at 12:26
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One key step in this problem is the fact that $U_{ab} \cong U_a \times U_b$, when $a$ and $b$ are co-prime. This comes from the Chinese Remainder Theorem.

The order of $U_a$ is $\phi(a)$ and the order of $U_b$ is $\phi(b)$. In most cases, both $\phi(a)$ and $\phi(b)$ are even and so $m=lcm(\phi(a),\phi(b))<\phi(a)\phi(b)=\phi(ab)$ is an exponent for $U_{ab}$, in the sense that $x^m =1$ for all $x\in U_{ab}$. This proves that $U_{ab}$ is not cyclic.

The remaining cases, when $n$ cannot be written as $n=ab$ with $a,b$ co-prime and $\phi(a),\phi(b)$ both even, give you the direction to the solution.

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  • $\begingroup$ Instead of gcd, you mean lcm, I think. $\endgroup$ – Saaqib Mahmood Apr 2 '14 at 11:58
  • $\begingroup$ But $U_{18}$ IS cyclic although $18 = 2 \times 9$ and gcd($2, \ , 9$) $=1$. $\endgroup$ – Saaqib Mahmood Apr 2 '14 at 12:00
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    $\begingroup$ @SaaqibMahmuud, thanks for the correction. In the example you gave, $\phi(2)$ is not even. $\endgroup$ – lhf Apr 2 '14 at 12:01
  • $\begingroup$ Ihf, can you please write out a complete proof for me? I'm rather rusty on even the elementary number theory! $\endgroup$ – Saaqib Mahmood Apr 2 '14 at 12:07
  • $\begingroup$ @Saaqib, there are proofs in nearly every introductory number theory textbook ever published, and probably on a thousand websites. Why should we duplicate all that effort here? Just go to a library, or type "primitive root" into Google. $\endgroup$ – Gerry Myerson Apr 2 '14 at 12:24

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