I am trying to prove the following elementary property (which I guess should be true):
Given 2 coprime positive integers $a, b$, we have $a\mathbb Z/b\mathbb Z = \mathbb Z/b\mathbb Z.$
Is the following proof correct? Is there a much faster way to show it?
I know that $(a, b)=1$ implies that there are integers $u$ and $v$ such that $ua+vb = 1$. In particular $ua =1 \bmod b$. Now essentially we need to show that $$\{a\bmod b, 2a\bmod b, \dots, ba\bmod b\}=\{0,1,\dots, b-1\}\,.$$ This will be true if the LHS has cardinality $b$. So if it does not hold there must be two integers $i\neq j$ in $\{0,1,\dots, b-1\}$ such that $ia = ja\bmod b$. In particular $(i-j)a=0\bmod b$. But then $0 = (i-j)ua = i-j\bmod b$, meaning that $i=j$ as they are both in $\{0,1,\dots, b-1\}$. This is a contradiction.