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n divides the totient of $p^n-1$ when p is prime

Math 2年 前

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.15
Let $p$ be a prime and let $n$ be a positive integer. Find the order of $p$ in $G = (\mathbb{Z}/(p^n-1))^\times$ and deduce that $n|\varphi(p^n-1)$, where $\varphi$ denotes the Euler totient function.


Solution: We have $p^n-1 \equiv 0$ in $G$, hence $p^n = 1$. Thus $|p|$ divides $n$.
Moreover, if $1 \leq k < n$, then $p^k < p^n-1$, so that $p^k \not\equiv 1$. Thus $|p| = n$.
By Lagrange's Theorem we deduce that $n|\varphi(p^n-1)$.

#Euler Totient#Lagrange’s Theorem#Prime Number
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