Question

Prove that a^m is a generator of G if and only if (m,n) = 1.

Short answer

Answer

It is proved that a^m is a generator of G if only if (m,n) = 1.

Step-by-step solution

Step by Step Solution Step 1: Required Theorem

Theorem 1.2: Let us consider two integers a and b, both not 0. Let d be their greatest common divisor. Then, there exists (not necessarily unique) integer u and v such that d = au + bv.

Prove that am is a generator of G if only if  (m,n) = 1.

From part (a) , we know that:

$〈a^m〉=〈a^d〉$

$$\begin{gathered} Sinced\in \left(m,n\right),take\left(m,n\right)=1. \\ Weget,〈a^m〉=〈a〉=G. \end{gathered}$$

From this, a^m is a generator of G, but if a^m is a generator of G, there should be some $u,v\in \text{Z}$, such that mu+nv=1.

Therefore, if d is the common divisor of both m and n, then it must also divide 1, which implies (m,n)=1 .