Let $l \geq 1$ be an integer and consider the cyclic group $(\mathbb{Z}/l\mathbb{Z},+)$. Given $n \in \mathbb{Z}$, we write $\bar{n} = n + l \mathbb{Z}$ for the class of $n$ in $\mathbb{Z}/l\mathbb{Z}$.
I would like to show that there is a well-defined composition law $\times$ on $\mathbb{Z}/l\mathbb{Z}$ such that $\bar{m} \times \bar{n} = \overline{m\times n}$ for all $m,n \in \mathbb{Z}$.
$\underline{Attempt:}$
Let the composition law be the additive operation $+$.
Let $\bar{m},\bar{n} \in \mathbb{Z}/l\mathbb{Z}$. Then we have the following:
$\bar{m}+\bar{n} = m + l \mathbb{Z} + n + l \mathbb{Z} = m+n+l \mathbb{Z} = \overline{m+n}$.
$\underline{Question}$
Could it be that this works with the composition law multiplication as well?
Let $\bar{m},\bar{n} \in \mathbb{Z}/l\mathbb{Z}$. Then we have the following:
$\bar{m}*\bar{n} = (m + l \mathbb{Z}) * (n + l \mathbb{Z}) = mn+ml\mathbb{Z}+nl\mathbb{Z}+(l\mathbb{Z})^2 =mn+l\mathbb{Z} = \overline{mn}$.
Would both ways be possible?
Answer to the question of user Gauss: When is the multiplication - defined in the second way - a group operation on $\mathbb{Z}/n\mathbb{Z}$?
I tried to think of it with the example $(\mathbb{Z}/5\mathbb{Z}, *)$. We have: $\mathbb{Z}/5\mathbb{Z}=\{\overline{0}, \overline{1}, \overline{2}, \overline{3}, \overline{4}\}$. Now closure under multiplication is given, since - as shown in my attemp - for any two elements they will in another class (for example $\overline{3} * \overline{4} = \overline{12} = \overline{2}).$ Associativity and neutral element ($\overline{1}$) shouldn't be a problem. What remains is to think about the inverse of such an element. Is that the correct way?
Inverse
$\overline{n} \in \mathbb{Z}/l\mathbb{Z}$ is invertible if and only if $gcd(n,l) =1$.
1.) Assume $gcd(n,l) =1$. By Bézouts Lemma we get that $\exists s,z \in \mathbb{Z}$ such that $s*n +t*l = 1 \iff sn = 1-t*l \iff sn $ mod $l = (1-t*l)$ mod $l \iff sn$ mod $l = 1 $ mod $l \iff s$ is the inverse element.
2.) Assume $\overline{n}$ is invertible $\iff n * n' + l\mathbb{Z} = 1 + l\mathbb{Z} \iff nn'$ mod $l = 1$ mod $l$, now ther exists a $t \in \mathbb{Z}$ such that $nn'$ mod $l = (1-t*l)$ mod $l \iff nn' = 1 - t*l \iff nn' + t*l = 1 \iff gcd(n,l)|1$. The last equivalence follows from Bézouts lemma.