show numbers (mod $p$) are distict and nonzero Let's start with the nonzero numbers, mod $p$,
$1$, $2$, $\cdots$, $(p-1)$,
and multiply them all by a nonzero $a$ (mod $p$).
Notice that if we multiply again by the inverse of $a$ (mod $p$) we get back the numbers 
$1$, $2$, $\cdots$, $(p-1)$.
But my question is how the above process show that the numbers
$a\cdot 1$ mod $p$, $a\cdot 2$ mod $p$, $\cdots$, $a\cdot (p-1)$ mod $p$
are distinct and nonzero?
Thanks in advance. 
 A: Suppose that $1\le x\lt y\le p-1$. 
If $ax\equiv ay\pmod{p}$ then $p$ divides $a(y-x)$. But $a$ and $p$ are relatively prime. So $p$ divides $y-x$. This is impossible, since $1\le y-x\lt p$.
None of the $ax$ is congruent to $0$ modulo $p$. For if $p$ divides $ax$, then $p$ divides $a$ or $p$ divides $x$. But $p$ does not divide $a$ by the choice of $a$. And $p$ does not divide $x$ since $1\le x\le p-1$.
Remark: Or else one can use multiplication by $a^{-1}$. The $ax$ must be distinct modulo $p$. For if $ax\equiv ay\pmod{p}$ then multiplying by $a^{-1}$ we find that $x\equiv y\pmod{p}$.  Since $x$ and $y$ are between $1$ and $p-1$, it follows that $x=y$.
It is easy to verify that none of the $ax$ is congruent to $0$ modulo $p$, else multiplying by the inverse would yield $0$.
So the $ax$ are distinct non-zero modulo $p$. There are $p-1$ of them. There are also $p-1$ possible values of $x$. So the $ax$ must be, in some order, congruent to all of the objects $1,2,\dots,p-1$. 
A: It is important to point out that $\Bbb Z_p$ is  a field if (and only if) $p$ is prime. This means in particular that every element that is not $0$ has an inverse. Since $p\not\mid a$ is equivalent to $a\not\equiv 0\mod p$, $a$ has an inverse $a^{-1}$. But then $$x\equiv y\mod p\iff ax\equiv ay\mod p$$
since we can reverse the equalities by multiplication by $a$ or $a^{-1}$. Moreover, since $p\not\mid a$ and $p\not\mid x$ (by assumption), $p\not\mid ax$, that is $$a\not\equiv 0,x\not\equiv 0\implies ax\not\equiv 0$$
A: The set you are considering is really a field: $\mathbb{Z}/p\mathbb{Z}$. Since it is a field, it is closed under multiplication, therefore when you multipliy them by a nonzero element $a$, all of them are distinct and nonzero (but they are somehow permuted)
