Prove that the non-zero integers modulo $p$, $p$ a prime number, form a group under multiplication $mod$ $p$, using the following criterion:

Criterion: Supose a finite set $G$ is closed under an associative product and that both cancellation laws hold in $G$. Then $G$ is a group.

I'm stuck following this theorem's proof, particularly when proving that $\frac{\mathbb{Z}}{p\mathbb{Z}}-\{\bar{0}\}$'s multiplication is left-cancellative.

The proof starts by considering that $\bar{a}\bar{b}=\bar{a}\bar{c}$, so, by definition, $\bar{ab}=\bar{ac}$ which, in turn, means that there exists $k\in\mathbb{Z}$ such that $ab=ac+kp$, this is $pk=a(b-c)$, thus $p|a(b-c)$.

$p$ is a prime number, which is enough to conclude that $p|a$ or $p|(b-c)$.

<What I don't understand starts here>: We have that $p\not|a$, $p\not|b$ and $p\not|c$ (I don't understand why this three truths hold), so $p|(b-c)$ (I get this), which in turn means that $b-c=0$ (I don't get this) and $b=c$.

I'd be really grateful if you explained why those two parts I don't get are true. I suppose they are obvious, so that's why the proof doesn't go any further in their development, but this is my first course in Group Theory and Number Theory, so maybe it's not that obvious for me.

Thanks in advance.

  • 4
    $\begingroup$ $a, b$, and $c$ are taken to be elements of $\Bbb Z/p\Bbb Z - \lbrace 0 \rbrace$. If $p \mid a$, for instance, then $a \equiv 0 \bmod p$, so $p \nmid a$ (else $a \notin \Bbb Z/p\Bbb Z - \lbrace 0 \rbrace$). Then, by the preceding conclusion, you have $p \mid b - c$, i.e. $b - c \equiv 0 \bmod p$, or alternatively, that $b \equiv c \bmod p$, so $b = c$ in $\Bbb Z/p\Bbb Z - \lbrace 0 \rbrace$. $\endgroup$ Nov 18, 2020 at 7:25
  • $\begingroup$ @EdwardEvans 's answer hit the nail. $\endgroup$ Nov 18, 2020 at 7:52


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