Multiplicative group of a finite prime field is cyclic

Statement

In the language of modular arithmetic

Let $p$ be a prime number. Then, the multiplicative group modulo $p$ is a cyclic group of order $p-1$. In other words, it is isomorphic to the group of integers modulo $p-1$.

A generator for this multiplicative group is termed a primitive root modulo $p$. While the theorem states that primitive roots exist, there is no procedure or formula known for obtaining a primitive root.

In the language of fields

The multiplicative group of a prime field is a cyclic group.

Related facts

For fields

• Multiplicative group of a field implies at most n elements of order dividing n
• At most n elements of order dividing n implies every finite subgroup is cyclic
• Multiplicative group of a field implies every finite subgroup is cyclic
• Multiplicative group of a finite field is cyclic
• Classification of fields whose multiplicative group is uniquely divisible
• Classification of fields whose multiplicative group is locally cyclic

For commutative rings

• Classification of natural numbers for which the multiplicative group is cyclic

Noncommutative analogues

• Multiplicative group of a skew field implies every finite abelian subgroup is cyclic

Examples

The prime 2

The multiplicative group modulo $2$ is the trivial group, so this is not an interesting case.

The prime 3

The multiplicative group modulo $3$ is of order two, and the element $2$ is a primitive root in this case.

The prime 5

The multiplicative group modulo $5$ is of order four. The element $2$ is a primitive root. The powers of $2$ include all elements: $2^{0}=1,2^{1}=2,2^{2}=4,2^{3}=3$.

$3$ is also a primitive root. Its powers are $3^{0}=1,3^{1}=3,3^{2}=4,3^{3}=2$.

The prime 7

The multiplicative group modulo $7$ is of order six. $2$ is not a primitive root: it has order $3$, and its powers only include $1,2,4$. On the other hand, $3$ is a primitive root, with $3^{0}=1,3^{1}=3,3^{2}=2,3^{3}=6,3^{4}=4,3^{5}=5$.

Facts used

1. Multiplicative group of a field implies every finite subgroup is cyclic

Proof

The proof follows directly from fact (1).