Let $m,N\in\mathbb{N}_2$, and consider modulo $mN$. Let $u\in\{1,\dots,mN-1\}$ such that gcd$(mN,u)=1$. I want to show that the following two sets are equal, where order is unimportant; $$uN\{0,1,2,\dots,m-1\}\equiv N\{0,1,2,\dots,m-1\}\pmod{mN}.$$ (For completion, we take $cA\pmod{n}\equiv\{ca\pmod{n}:a\in A\}$)
For example; let $m=6$, $N=24$, and let $u=5$, which satisfies gcd$(mN,u)=$gcd(144,5)$=1$. Then we have \begin{equation} \begin{split} 5\times 24\{0,1,2,3,4,5\}\equiv 5\{0,24,48,72,96,120\} \equiv&\: \{0, 120, 240,360,480,600\}\\ \equiv&\: \{0,120,96,72, 48, 24\} \equiv 24\{0,1,2,3,4,5\}\pmod{144}. \end{split} \end{equation}
I've been struggling with this for a few days now. It appears simple, but I just keep getting lost in knots of logic. I could also just be missing something obvious... Any help is appreciated.
Treating all the numbers as elements of $\mathbb{Z}/mN\mathbb{Z}$ is more convenience. The set $S=N\{0,1,2,\dots,m-1\}$ is the set of all multiples of $N$ in $\mathbb{Z}/mN\mathbb{Z}.$ Let the map $f(x) = ux$ be the function from $S$ to its image $uS = uN\{0,1,2,\dots,m-1\}.$ The image of $f$ is a subset of $S$ since if $x$ is a multiple of $N,$ then so is $ux,$ therefore $uS \subset S.$ $f$ is bijective because it is invertible with the map $x \mapsto u^{-1}x,$ so $S$ and $uS$ has the same size, and therefore they are the same.