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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?

MyNameIs
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  • 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. – Piquito Jul 24 '16 at 14:56
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    @Piquito: $(\mathbb{Z}/n\mathbb{Z})^*$ is a group, not a ring; it has no zero divisors. – anomaly Jul 24 '16 at 14:57
  • @Piquito what about $(\mathbb Z/4\mathbb Z)^* \cong \mathbb Z/2\mathbb Z$ ? – cat Jul 24 '16 at 15:17
  • @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, $;k\in\Bbb N;$ – DonAntonio Jul 24 '16 at 15:25
  • How hard is that to prove, @DonAntonio ? Do you care to sketch a proof for me? Could you give me a reference? – MyNameIs Jul 24 '16 at 15:59
  • @MyNameIs Try this one: http://people.reed.edu/~jerry/361/lectures/lec07.pdf There are thousands of different sources in the web and also in books. – DonAntonio Jul 24 '16 at 16:23

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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.

P Vanchinathan
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