Let's say I have $M-1$ integers, all of them different from each other, and all of them smaller than integer M: $$1,2,3...M-1$$ I multiply each of them by another integer S, and write the result modulo M . So, I get another group of numbers :$$r_1,r_2...$$ I would like to prove that the new group that I found is permutation of the ordinary group:$$1,2,3...M-1$$
Could someone help? Thanks.
If $(M,S)\neq 1$ then this result is false. (Take any two not coprime numbers and you'll see).
Assuming $(M,S)=1$ then consider the set $\mathcal{S}=\{S\cdot k\mod M\mid k=0,\ldots,M-1\}$.
Proving that $\mathcal{S}$ is the original set (disregarding order) is equivalent to proving that if $Sa\equiv Sb\pmod{M}$ with $a,b$ from the original set, then $a\equiv b\pmod{M}$.
If $Sa \equiv Sb$ then $Sa-Sb\equiv 0\Rightarrow S(a-b)\equiv 0 \Rightarrow M\mid S(a-b)$. Since $(M,S) = 1$ then $M\mid (a-b) \Rightarrow a\equiv b\pmod {M}$.