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Let $p$ be a prime. Suppose that $a \in \Bbb Z$ is not divisible by $p$. Show that the congruence class $[a]_p$ has an inverse with respect to multiplication of congruence classes.

I am not very familiar with congruence classes, but it looks like $[a]_p = \{b \in \Bbb Z \mid b \equiv a \pmod{p} \}$. What I need to show is that there exists $a^{-1}$ such that $[a]_p \cdot [a^{-1}]_p = [1]_p$?

Only thing I can get out of this is that if $p$ doesn't divide $a$, then $a = pk + r$ where $k$ is some integer and $0<r\le k$ a remainder which is just another way of saying that $a \equiv r \pmod{p}$, where $r \ne 0$.

How is this related to the fact that there should exist an inverse for $[a]_p$?

Zwertp
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  • Hint : $\ \mathbb Z_p\ $ is a field if $\ p\ $ is prime. Show this and you are done. You can also use Bezout's identity and the fact that every number $1,\cdots,p-1$ is coprime to the prime $p$. – Peter Nov 11 '21 at 13:01
  • I suppose we're talking about $\Bbb Z_p$ as the group with multiplication not addition? The group is then defined as the elements from $1, \dots, p-1$ such that every element is coprime to $p$ right? This would just be all the elements starting from $1$ to $p-1$? @Peter – Zwertp Nov 11 '21 at 13:04
  • This is a possible way to interprete it. Note that $\mathbb Z_p$ is a finite integral domain (it is easy to show that it has no zero-divisor) and therefore every non-zero element is a unit and therefore has a multiplicative inverse. – Peter Nov 11 '21 at 13:14

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