Primitive roots and 'equivalent exponents'.

Question

If M is a primitive root mod p and M = $\ N^T$ mod p , then the order of N mod p is also (p-1) is this true?

Answer 1

Yes, it is true. For suppose that $N$ has order $d\lt p-1$. Then $M^d=(N^T)^d=(N^d)^T\equiv 1\pmod{p}$. So $M$ has order $\le d$, contradicting the fact that $M$ is a primitive root of $p$.

Remark: More informally, since $M$ is a power of $N$, and the powers of $M$ range (modulo $p$) over the set $\{1,2,3,\dots,p-1\}$, so do the powers of $N$. "Everything" is a power of $N$, so $N$ is a primitive root of $p$.