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So I encountered this proof on a Number Theory book, I will link the pdf at the end of the post (proof at page 96), it says: "Every prime has a primitive root, proof: Let p be a prime and let m be a positive integer such that: p−1=mk for some integer k. Let F(m) be the number of positive integers of order m modulo p that are less than p. The order modulo p of an integer not divisible by p divides p − 1 , it follows that: $$p-1=\sum_{m|p-1}F(m) $$ By theorem 42 we know that: $$p-1=\sum_{m|p-1}\phi(m) $$ By Lemma 11,F(m)≤φ(m) when m|(p−1). Together with: $$\sum_{m|p-1}F(m)=\sum_{m|p-1}\phi(m) $$ we see that F(m)=φ(m) for each positive divisor m of p−1. Thus we conclude that F(m)=φ(m). As a result, we see that there are p−1 incongruent integers of order p−1 modulo p. Thus p has φ(p−1) primitive roots.

The part that i don't understand is near the beginning, when he says "The order modulo p of an integer not divisible by p divides p − 1 , it follows that: $$p-1=\sum_{m|p-1}F(m) $$" How does he conclude that? I understand that the order of the integer must divide p-1 but how does that imply that the summation actually evaluates at p-1?...

Link of the book's pdf: https://www.saylor.org/site/wp-content/uploads/2013/05/An-Introductory-in-Elementary-Number-Theory.pdf

Bill Dubuque
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Spasoje Durovic
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  • Try this for some small prime like $5$ or $7$ and you'll see. – saulspatz Jul 23 '18 at 13:57
  • The nonzero elements of the $p$-element field form a group under multiplication of order $p-1$; thus the multiplicative order must be a divisor of $p-1$. – egreg Jul 23 '18 at 14:10
  • Recall how you defined $F(m)$, "Let $F(m)$ be the number of positive integers of order $m\bmod p$ that are less than $p$." What's the order of the group? – sharding4 Jul 23 '18 at 14:29
  • Nice proof actually. Another proof is a corollary of the (much longer) result that if $(F, +,0,\times, 1)$ is a finite field then the group $(F\backslash {0}, \times, 1)$ is cyclic. – DanielWainfleet Jul 23 '18 at 15:03

2 Answers2

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There are $p-1$ positive integers less than $p$, namely $1, 2, ..., p-1$.

Each of these will have some multiplicative order modulo $p$. So if we count all those of order $1$, all those of order $2$, all those of order $3$, etc then the total count is $p-1$. There are $F(1)$ of order $1$, $F(2)$ of order $2$, etc, so:

$$p-1=\sum_{m=1}^{\infty}F(m)$$

However, we know their orders will divide $p-1$, so almost all the terms in this sum will be zero. Only those with $m|(p-1)$ will contribute to the sum. We therefore have:

$$p-1=\sum_{m|p-1}F(m)$$

Jaap Scherphuis
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All that is saying is that the sum of the number of elements of each order must add up to the total number of elements in the field, p -1. The summation could have been carried out of "all possible orders," but since "all possible orders" is the same thing as "divisors of p -1," we can carry the summation over divisors of p -1. The summation equation is literally equivalent to the statement that the order of a member of the field must be a factor of p - 1

Tom Healy
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