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Definition: $(\mathbb{Z}/n \mathbb{Z})^{\times} = \{\bar{a} \in \mathbb{Z}/n \mathbb{Z}: \gcd(a, n) = 1\}$.

I know that $(\mathbb{Z}/9 \mathbb{Z})^{\times}$ and $(\mathbb{Z}/6 \mathbb{Z}))^{\times}$ are cyclic because $\langle\bar{5}\rangle = (\mathbb{Z}/9 \mathbb{Z})^{\times}$ and $ \langle\bar{2}\rangle = (\mathbb{Z}/6 \mathbb{Z})^{\times}$, but $(\mathbb{Z}/8 \mathbb{Z})^{\times}$ is not cyclic. Is there a general rule to say if $(\mathbb{Z}/n \mathbb{Z})^{\times}$ is cyclic?

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Yes: $(\mathbb Z/n\mathbb Z)^\times$ is cyclic if and only if $n$ is either an odd-prime power, or $2p^n$ with odd prime $p$.

One direction of the proof is relatively easy, since $\mathbb Z/p^mq^n\mathbb Z\approx \mathbb Z/p^m\mathbb Z \oplus \mathbb Z/q^n\mathbb Z$... making it hard for the multiplicative group to be cyclic.

The prime-power case is slightly subtler, and the factor-of-$2$ aspect also.

EDIT: and, as @lhf reminded me, for $m=1,2,4$! :)

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  • $\begingroup$ Also, $n=1,2,4$. $\endgroup$
    – lhf
    Sep 27 '21 at 23:38

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