solving congruence class equation

Question

Keeping $p = 7$ is the following true?

If for any congruence class $[a]$ that is not $[0]$, there is a unique class $[b]$ such that $[a] \cdot [b] = [1]$.

Would this be true? (assuming you keep $p = 7$)

Answer 1

One can show that $\mathbb{Z}/p\mathbb{Z}$ is a field if and only if $p$ is a prime. This implies that every element has a multiplicative inverse in this group.

However, if $p$ is not a prime, you have a ring with zerodivisors (and hence are not fields): take $x \in \mathbb{Z}/n\mathbb{Z}$ such that $x \mid n$, which is possible since $n$ is not prime (in particular $1 < x < n$. This implies that $n = dx$ for some integer $d$ and hence we have that $[x][d] = [n] = 0$, whereas $x \neq 0$ nor $d \neq 0$. Hence these elements can't have multiplicative inverses.

Answer 2

If there are two classes $[a]$ and $[b]$ such that $[a]\cdot [c]=[b]\cdot [c]=1$ that is, $[a]$ and $[b]$ are the inverse of $c$.

Since $\Bbb Z_7^*$ is a group, we can use the cancellation property and prove that $[a] = [b]$.