Example

If g is an element of a cyclic group G of order n, then the order of g is a divisor of n.

We can use this to show that 2 generates the multiplicative group of Z/101Z:

Since 101 is prime, the group (Z/101Z)^* of invertible elements of Z/101Z contains 100 elements. By Fermat's little theorem we also have that 2¹⁰⁰ = 1 mod 101. Thus the order of 2 is a divisor of 100. Easy computations show that

• 2¹⁰ = 14 mod 101,
• 2²⁰ = -6 mod 101, and
• 2⁵⁰ = -6 · (-6) · 14 = 504 = -1 mod 101.

Hence the order of 2 is neither a divisor of 50 nor of 20, and so it is 100.