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Let $m \in \mathbb{N}$. I want to show that there's a bijection between the sets $$A = \{(q,a) \mid q\in \mathbb{N} \text{ divides }m, \text{ and }a\in \mathbb{Z}/q\mathbb{Z} \text{ is such that }\gcd{(a,q)}= 1\}, \text{ and}$$

$$B = \mathbb{Z}/ m\mathbb{Z}.$$

For example, when $m = 6$,

$A = \{(1,1), (2,1), (3,1), (3,2), (6,1), (6,5)\}$.

I'm having difficulty finding an appropriate map between the two sets.

24601
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2 Answers2

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Hint: The set $A$ has $$ \sum_{d\mid m}\varphi(d)=m $$ elements. For example, for $m=6$ the divisors are $d=1,2,3,6$ and we have $\varphi(1)+\varphi(2)+\varphi(3)+\phi(6)=1+1+2+2=6$.

Reference: Summation involving totient function: $\sum_{d\mid n} \varphi(d)=n$

Dietrich Burde
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Let's rewrite your definition of $A$ as $$ A=\{(q,a):q\mid m,\quad 1\le a\le q,\quad\gcd(a,q)=1\} $$ where all variables are in $\mathbb{N}$.

Take $a$ such that $0\le a<m$ and consider $q=m/\!\gcd(a,m)$. Then $\gcd(a,q)=1$, so $(q,a)\in A$.

Conversely, if $(q,a)\in A$, you have $\gcd(a,m)=m/q$.

So the set $A$ essentially lists the integers in $[0,m)$ and the required map is $f(q,a)=am/q$.

Let's make an example with $m=6$. Then we have $$ A=\{(1,0),(2,1),(3,1),(3,2),(6,1),(6,5)\} $$ and we get $f(1,0)=0$, $f(2,1)=3$, $f(3,1)=2$, $f(3,2)=4$, $f(6,1)=1$, $f(6,5)=5$.

egreg
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