Classification of natural numbers for which the multiplicative group is cyclic

This article states and (possibly) proves a fact that is true for certain kinds of odd-order groups. The analogous statement is not in general true for groups of even order.
View other such facts for finite groups

Statement

Let $n$ be a natural number greater than $1$. Then, the Multiplicative group modulo n (?) is a cyclic group if and only if $n$ is $2,4$ or of the form $p^{k}$ or $2p^{k}$ for some odd prime $p$.

Caution

Note that the multiplicative group modulo $p^{k}$ is a totally different concept from the multiplicative group of a finite field of order $p^{k}$. The former arises as the multiplicative group of a ring that is far from a field, while the latter is the multiplicative group of a finite field. When $k=1$, the notions coincide.

Also, the multiplicative group of a finite field is always cyclic, even when the prime is $2$.

Related facts

• Multiplicative group of a prime field is cyclic
• Multiplicative group of a finite field is cyclic
• Multiplicative group of a field implies every finite subgroup is cyclic
• Classification of fields whose multiplicative group is locally cyclic

Examples

For examples of prime numbers, refer Multiplicative group of a prime field is cyclic#Examples.

Here, we discuss examples of prime powers.

$n=2^{2}=4$

In this case, the multiplicative group is $\{1,3\}$, and is cyclic on $3$.

$n=3^{2}=9$

In this case, the multiplicative group is $\{1,2,4,5,7,8\}$, and it has two possible generators: $2$ and $5$.