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$U(pq)$ is to be the units of $Z_{pq}$. I would assume that distinct prime factors would enable $U(pq)$ to always by cyclic. My professor alluded to the fact that this may not always be true. However, I am curious if this is always the case, since $U(p)$ is cyclic, given $p$ is prime.

What are your thoughts?

Shaun
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Ron Snow
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1 Answers1

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If $U(15)=U(3×5)$ were cyclic, there would be a primitive root modulo $15$, but there is none. Thus $U(15)$ is not cyclic.

$U(n)$ is cyclic iff $n=1,2,4,p^k,2p^k$ where $p$ is an odd prime and $k\ge1$.

Parcly Taxel
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  • So is there a way to prove that $U(n)$ is always cyclic if we can write $n$ as $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ with $k$ distinct primes? Or is this, by your argument, also never cyclic? – Ron Snow Oct 17 '19 at 16:42
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    @Edison See the duplicate. – Parcly Taxel Oct 17 '19 at 16:56