$p$ is prime and $a\in \mathbb Z$ Prove that this set is a complete residue system $\bmod p$

Question

if $p$ is prime and $a\in \mathbb Z$ such that $p\nmid a$, Show that the set $$ R=\{0,a,2a,3a,...,(p-1)a\}$$

Is a complete residue system modulo $p$.

This problem is from my textbook.

My attempt:

I think that the best way to prove it it's by contradiction, so let $x,y \in R$ such that : $$x\ne y, x\equiv y \pmod p$$

and we gonna prove that it's impossible that $x,y\in R$ and $x\equiv y \pmod p$... But I don't know how to do it.

score 1 · Accepted Answer · answered Feb 24 '21 at 15:38

1

Yes, $a$ is missing. Let $x=k_1a,y=k_2a$ where $k_i\in\{0,1,2,...,p-1\}$. Then $k_1-k_2\in\{-(p-1),...,-1,0,1,2,...,p-1\}$.

$p|(x-y)\implies p|(k_1-k_2)a$ and since $p\nmid a$, it must be that $p|(k_1-k_2)$. So $k_1-k_2=0$. Thus $x=y$.

answered Feb 24 '21 at 15:38

Shubham Johri

why $k_1-k_2 = 0$ – PNT Feb 24 '21 at 15:44
Because $0$ is the only value in ${-(p-1),...,-1,0,1,...,p-1}$ that is divisible by $p$. – Shubham Johri Feb 24 '21 at 15:45
and why $p\mid x-y$ – PNT Feb 24 '21 at 15:46
Well you assumed $x\equiv y\mod p$. – Shubham Johri Feb 24 '21 at 15:49
1

yeah I forgot, thank you – PNT Feb 24 '21 at 15:50

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