I want to prove $\overline{ab} =\overline{a}\cdot\overline{b}$. Here $\overline{a} $ and $\overline{b}$ are elements of $\Bbb Z_n$. I am really having a bad time to prove it. I needed it to prove that $(\Bbb Z_n,\cdot)$ is a Semigroup.
My Try : $\overline{a}\cdot \overline{b} $ is a subset of $\overline{ab}$. But I can not prove the other way around. Can you please help me by giving a hint?
Thank You in Advance.
You can't prove it the other way around, because it's not true. For instance, $\overline n.\overline n$ as set operation in $\Bbb Z$ doesn't contain $n$ (as long as $|n|>1$). However, that doesn't matter because that's not how $\bar a.\bar b$ is defined in $\Bbb Z_n$.
In $\Bbb Z_n$, $\bar a.\bar b$ is defined as $\overline{ab}$. The only thing we have to do is what you've already done: show that the resulting congruence class doesn't depend on what representative we choose for $\bar a$ and $\bar b$. In other words, that the $\Bbb Z$-product $\bar a.\bar b$ is a subset of $\overline{ab}$.