$\mathbb Z_2$ has no zero divisors while $\mathbb Z_6$ has zero divisors? Zero divisor is defined as an element $a\not = 0$ of a ring $R$ if $\exists b\not = 0$ such that $ab=0$ or $ba=0$.
Now, $2$ is not a zero divisor of $\mathbb Z_2$ since it is equivalent to the coset $0$, more here. But $2$ is a zero divisor of $\mathbb Z_6$ but I cannot understand why.
Please demonstrate why $\mathbb Z_6$ has zero divisors while $\mathbb Z_2$ does not have zero divisors: I cannot fully understand the notation $a\not = 0$ to mean coset and by this question I am trying to understand it better.
How can you intuitively demonstrate that $\mathbb Z_2$ has no zero divisors while $\mathbb Z_6$ has zero divisors 2 and 3?
Thank you in advance.
 A: There are only two elements in $\Bbb{Z}_2$, $[0]$ and $[1]$. As you said, in $\Bbb{Z}_2$, $[2]=[0]$, so by definition it is not a zero divisor. The only other option is $[1]$. But $[1]\cdot[1]$ is not $[0]$, and we can't use $[1]\cdot[0]=[0]$ according to the definition, so $\Bbb{Z}_2$ has no zero divisors.
However, in $\Bbb{Z}_6$, we have $[2]\cdot[3]=[6]=[0]$, so both $[2]$ and $[3]$ are zero divisors.
The best intuitive explanation I have is that $\Bbb{Z}_2$ and $\Bbb{Z}_6$ are completely different groups. In particular, $[2]$ in $\Bbb{Z}_2$ is NOT the sam as $[2]$ in $\Bbb{Z}_6$. In the first case, $[2]=\{\ldots, -2, 0, 2, 4,\ldots\}$. In the second, $[2]=\{\ldots-4, 2, 8, 14\ldots\}$. That we use the same notation for both may be a source of confusion.

Here, $[a]$ indicates an equivalence class. In $\Bbb{Z}_n$, $[a]=\{a+kn:k\in\Bbb{Z}\}$. If some of this is new to you, I recommend you do a quick google search of modular arithmetic.
A: Within modular arithmetic, $\forall a \in \mathbb Z_n , \forall b \in \mathbb I$ an = 0
Substituting n for 0 in in the original definition, we get that a is a zero divisor when ab=n or ba=n.
This is true if and only if a is a factor of n.
Looking at the factorization of 2, {2}, one can intuit that 2 is the only zero divisor in $\mathbb Z_2$. 
Likewise, the factorization of 6, {2,3,6}, reveals that in addition to 6, 2 and 3 also serve as zero divisors in $\mathbb Z_6$.
(Note: We exclude 1 from the factorization).
