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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?

Saaqib Mahmood
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    Exactly when $;n=1,,,n=2,,,n=2p,,,p^k;$ , with $;p;$ an odd prime. – DonAntonio Apr 02 '14 at 11:05
  • This would have been better if you include "I think" with "which, when" rather with only "which,when"... –  Apr 02 '14 at 11:06
  • This is explained in almost all books on number theory. – lhf Apr 02 '14 at 11:06
  • Ihf, I'd appreciate if you could also give an exact reference where both parts of my question are answered. – Saaqib Mahmood Apr 02 '14 at 12:04
  • DonAntonio, please elaborate your answer with an easy-to-understand proof. I'm afraid I'm not nearly so gifted as you are! – Saaqib Mahmood Apr 02 '14 at 12:06
  • See http://math.stackexchange.com/questions/42755/order-of-cyclic-groups and my answer there. See also http://math.stackexchange.com/questions/314846/for-what-n-is-u-n-is-cyclic. – lhf Apr 02 '14 at 12:07
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    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. – Gerry Myerson Apr 02 '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.

lhf
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