Are there exactly $N$ different irreductible fractions less or equal than $1$ with denominator divisor of $N$?

Question

Solving an exercise, I saw some cases that this was true, like $N=6$, where the $6$ fractions are $1, 1/2, 1/3, 2/3, 1/6, 5/6$, and then intuited it was always true. I guess we can prove it writing the prime factorization of $N$: $a^Ab^B...k^K$, and using the numbers of divisors, $(A+1)(B+1)...(K+1)$.

Answer 1

Yes. All fractions in $(0,1]$ with denominators as divisors of $N$ must equal $k/N$ for some integer $1\leq k\leq N$, and each of those reduce to exactly one irreducible fraction.

Answer 2

Different viewpoint: the fractions with denominator $d$ that are already in lowest terms have numerator coprime to $d,$ and no larger than $d$ itself. The count of those numerators is the Euler totient $\phi(d)$

There is then a fairly easy number theory fact using "multiplicative" functions, $$ \sum_{d|n} \; \phi(d) = n $$

This identity has appeared many times as a question... Is there a direct, elementary proof of $n = \sum_{k|n} \phi(k)$?