Proving the Generators of a Cyclic Group and the Role of GCD

In the context of group theory, particularly in dealing with cyclic groups, understanding the conditions under which an element generates the entire cyclic group is crucial. Specifically, we will prove that if a cyclic group is generated by one element a of order n, then a^m is a generator of the group if and only if gcd(m,n) 1.

Understanding Cyclic Groups and Generators

A cyclic group is a group that is generated by a single element. This means that all elements of the group can be expressed as powers of that single generator element. The order of a generator a is the smallest positive integer n such that a^n e, where e is the identity element of the group.

Proof of the Theorem

Direction 1: If gcd(m,n) 1, then a^m generates the cyclic group

To prove the first direction, we assume gcd(m,n) 1. If a^m does not generate the cyclic group, there exists an integer k such that a^{mk} e. Given that the order of a is n, a^n e, we can conclude that n divides mk. However, because gcd(m,n) 1, n and m are coprime, meaning n cannot divide m. Thus, n must divide k. This contradicts the minimality of n as the order of a because k n but n divides k. Therefore, a^m must generate the cyclic group.

Direction 2: If a^m generates the cyclic group, then gcd(m,n) 1

Suppose a^m generates the cyclic group. Let d gcd(m,n). By Bezout's theorem, there are integers u and v such that dmnv 1. This gives us the equation:

a^d a^munv a^m^u?a^n^v a^m^u (since a^n e).

Since d divides n, n ds for some integer s. Thus, a^n a^ds and (a^d)^s e. Therefore, the order of a^d is a divisor of s. Since d divides m (as um and vn are integers), we have a^m (a^d)^u e. This implies that the subgroup generated by a^d is contained in the one generated by a^m.

Since a^m generates the whole group, a^m must generate the subgroup generated by a^d. Therefore, the order of a^m must be the same as the order of the group, which is n. Thus, n divides mk. Since d divides both m and n, and gcd(m,n) d, we must have d 1.

Conclusion

By proving both directions, we establish that a^m generates the cyclic group if and only if gcd(m,n) 1. This result highlights the importance of the greatest common divisor in determining the generators of a cyclic group.

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