Multiplicative group of a prime field is cyclic

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.

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

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) uses::Multiplicative group of a field implies every finite subgroup is cyclic

Proof
The proof follows directly from fact (1).