multiplication in $\mathbb{Z}_p$ I'm not even sure what would be a good title for my question, so feel free to edit it (if you can).
I want to prove (or dispute) the correctness of the following lemma: 

Let $p$ be a prime number, and let $i,j,c \in \mathbb{Z}_p \setminus
 \{0\}$ such that $i \neq j$. Is it true that $ic \neq jc \pmod p$?

In other words, what I want to show is that given any $c \in \mathbb{Z}_p \setminus \{0\}$, the results of $c, 2c, \cdots , (p-1)c$ are pairwise different.
First - is this even correct? a short python program I wrote for small prime numbers gives me the feeling it is.
Second - if it's correct -how do I prove it? I tried assuming that $ic = jc \pmod p$ but got stuck with it.
Thanks
 A: Recall that an integral domain is a commutative ring such that $ab=0$ implies $a=0$ or $b=0$.
The essential property of primes I'm going to use is the fact that if $p$ is prime and $p|ab$ then $p|a$ or $p|b$.

Lemma. $\mathbb{Z}_n$ is an integral domain if and only if $n$ is prime.

Proof. "$\Rightarrow$" assume $n$ is not prime and take $p|n$ a prime divisor. Then both $p, n/p\in\mathbb{Z}_n$ are non-zero but
$$p\cdot n/p=n=0\text{ (mod n)}$$
Contradiction.
"$\Leftarrow$" Let $ab=0 \text{ (mod p)}$ for prime $p$, i.e. $p|ab$. Since $p$ is prime then this implies that either $p|a$ or $p|b$, i.e. $a=0$ or $b=0$ in $\mathbb{Z}_p$. $\Box$

With that it is easy to prove the statement: assume that $i,j\in\mathbb{Z}_p$ and $c\in\mathbb{Z}_p\backslash\{0\}$ are such that
$$ic=jc\text{ (mod p)}$$
Then
$$ic-jc=0\text{ (mod p)}$$
$$(i-j)c=0\text{ (mod p)}$$
and since $\mathbb{Z}_p$ is an integral domain then $i-j=0\text{ (mod p})$ because $c\neq 0$. Therefore $i=j\text{ (mod p)}$ which completes the proof.
