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.

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).