Some time ago, I posted a question on six primes numbers forming an arithmetic progression, and someone commented the following: " Let $m$ and $n$ be positive integers with $\gcd(m,n)=1$. Then the sequence $m,2m,3m,…,nm$ when taken mod $n$ will give the integers from $1$ to $n$ , each occurring exactly once ". I wanted to know why so I tried to "prove" it, this what I did:
I supposed the contrary, that in the sequence $a_1 m \equiv a_2 m \pmod n$, with the condition that $a_1\not \equiv a_2 \pmod n$ (I do not know if it´s possible to use that notation but it just means that they are not congruent) , because $a_1 \neq a_2$ and because both $a_1$ and $a_2$ and have to be less or equal to $n$ .
So I did this:
$a_1 m \equiv a_2 m \pmod n$
$a_1 m - a_2 m \equiv 0 \pmod n$
$(a_1 - a_2)m\equiv 0 \pmod n$
This means that $(a_1 - a_2)m$ is a multiple of $n$, but this is a contradiction because $\gcd(m,n)=1$ and since $a_1\not \equiv a_2 \pmod n$ then $a_1 - a_2 \not \equiv 0 \pmod n$, which means that $a_1 - a_2$ doesn´t have all factors of $n$ and therefore it is not a multiple.
This may be in a textbook, so sorry if I made you lose your time.
Anyway, I would just like you to tell me if it´s ok what I did.