I am currently looking at the group $M:= (\mathbb{Z}/m\mathbb{Z})^* :=(\mathbb{Z}/m\mathbb{Z})\setminus m\mathbb{Z}$, the multiplicative group of integers modulo m.
I need to show that M is closed under multiplication for $m\geq 2$ iff m is prime.
Here we use that $ p \geq 2$ is prime iff with $p|kl$ then $p|k$ or $p|l$.
Why is it, that if $M$ is closed under multiplication, then for $k,l \notin m\mathbb{Z}$ also $kl\notin m\mathbb{Z}$?
Why is if $k,l \notin m\mathbb{Z}$ also $kl\notin m\mathbb{Z}$, then this means that m is prime?
My book says that the closure of M means that $ k \notin m\mathbb{Z}$ and $ l \notin m\mathbb{Z} \Rightarrow kl\notin m\mathbb{Z} $ and that this is equivalent to the definition of prime in the first question $$ p|kl \Rightarrow p|l \wedge p|k $$. I don't understand why the closure means that and why it is equivalent to the definition of prime. I can only understand the question about the closure if assumed that $k,l\in M$ but that makes no sense, because it would naturally mean that $k,l \notin m\mathbb{Z}$.