Chapter 5: Problem 7
Show that if \(a\) and \(n\) are relatively prime with \(n>0\), then ord \(_{n} a \mid \phi(n)\).
Short Answer
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Question: Prove that ord$_{n}a | \phi(n)$ for relatively prime numbers $a$ and $n$.
Answer: We can prove this result using properties of modular arithmetic and Euler's theorem. By definition, ord$_{n}a$ is the smallest positive integer $k$ such that $a^k \equiv 1 \pmod{n}$. Being relatively prime, Euler's theorem applies and we have $a^{\phi(n)} \equiv 1 \pmod{n}$. Let $d = $ ord$_{n}a$. Using the division algorithm, we can write $\phi(n) = pd + r$, where $0 \le r < d$. It can be shown that $a^r \equiv a^{\phi(n)} \pmod{n}$, leading to the conclusion that $r = 0$. This implies $\phi(n) = pd$, which means ord$_{n} a = d \mid \phi(n)$.
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Key Concepts
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