Modulo values of indivisible numbers

Question

Given is prime number $p$ and natural number $a$ which is relatively prime to $p$. Prove that no numbers from the set $B={0a,1a,2a,...,(p-1)a}$ give the same value after divison by $p$.

Answer 1

Suppose there are such numbers: $ma$ and $na$ ($n>m$), this would mean, that $p|a(n-m)$ since $p$ and $a$ are relative prime, this lead to $p|(n-m)$. But that is impossible, since $0<n-m<p$. Hence no such numbers exist and every element of the set gives different value$\pmod{p}$.

Answer 2

If $p$ divides $r_1a-r_2a=(r_1-r_2)$ where $p-1\ge r_1>r_2\ge1\ \ (1)$

We need $p(r_1-r_2)$

But by $(1),$ $$0<r_1-r_2<p-1$$ which is not divisible by $p$

Answer 3

What if they did?

Suppose that $ak \equiv aj \equiv m\pmod p$ for some $0 \le m < p$. What would that mean.

It would mean that $ak = m + p*w$ for some integer $w$ and then $aj = m + p*v$ for some $v$.

If we subtract those then $ak - aj = p*w -p*v$ so $a(k-j) = p(w-v)$

So $p|a(k-j)$.

Can you finish?