Let the generator of N-order cyclic multiplication group G be a, then a n = 1. G 1 is a subgroup of g.

K is G 1 element with the smallest exponent, then

(a k) * (a k) = a (2k) is still an element of G 1. If a k ≠ 1, then a (2k) ≠ ak;

By analogy, if a (2k) ≠ 1, then a (3k) ≠ a k, a (3k) ≠ a (2k),

……

So a k is the generator of G 1

The subgroup G 1 of ∴G is still a cyclic group.