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Is the group of units $(\Bbb Z/n\Bbb Z)^\times$ always cyclic? Do we need that $n$ is a prime or something?


marked as duplicate by anomaly, Strants, hardmath, awllower, Henning Makholm Jul 24 '16 at 15:00

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  • $\begingroup$ Because of $(\Bbb Z/n\Bbb Z)^*$ has divisor of zero if and only if $n$ is composed, this group is cyclic if and only if $n$ is prime. $\endgroup$ – Piquito Jul 24 '16 at 14:56
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    $\begingroup$ @Piquito: $(\mathbb{Z}/n\mathbb{Z})^*$ is a group, not a ring; it has no zero divisors. $\endgroup$ – anomaly Jul 24 '16 at 14:57
  • $\begingroup$ @Piquito what about $(\mathbb Z/4\mathbb Z)^* \cong \mathbb Z/2\mathbb Z$ ? $\endgroup$ – cat Jul 24 '16 at 15:17
  • $\begingroup$ @Piquito That is false. The group $\;\left(\Bbb Z/n\Bbb Z\right)^*\;$ is cyclic iff $\;n=1,2,4,p^k, 2p^k\;$ , with $\;p\;$ an odd prime, $\;k\in\Bbb N\;$ $\endgroup$ – DonAntonio Jul 24 '16 at 15:25
  • $\begingroup$ How hard is that to prove, @DonAntonio ? Do you care to sketch a proof for me? Could you give me a reference? $\endgroup$ – MyNameIs Jul 24 '16 at 15:59

It is cyclic for $n=4$ and for $n=p^k$ with $p$ an odd prime, $k\geq1$. (read books on number theory, e.g. Ireland and Rosen.)

For $n=8$ check that the unit group consists of $\{1,3,5,7\}$ with operation being multiplication mod $8$. Each number is its own inverse and hence no element of order 4 in this group (So this is isomorphic to Klein's group).

EDIT: This list is incomplete. See the comment of DonAntonio below.

  • $\begingroup$ ...and for $\;n=1,2, p^k\;$ , with $\;p\;$ an odd integer and $\;k\in\Bbb N\;$ . $\endgroup$ – DonAntonio Jul 24 '16 at 15:26
  • $\begingroup$ Thanks for adding the missing cases! $\endgroup$ – P Vanchinathan Jul 24 '16 at 15:34
  • $\begingroup$ That should have been $\;n=1,2,2p^k\;,\;\;k\in\Bbb N\;$ . My pleasure. $\endgroup$ – DonAntonio Jul 24 '16 at 15:38

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