Trying to understand quotient ring. From my understanding of the definition:
$(G/U,+,\cdot)$ is quotient ring, where the set $G/U=\{a + U \mid a \in G\}\stackrel{?}{=}\{a+b \mid a \in G, b \in U\}$.
For example:
$$\mathbb{Z}=\{\dots,-3,-2,-1,0,1,2,3,\dots\}$$
$$6\mathbb{Z}=\{\dots,-18,-12,-6,0,6,12,18,\dots\}$$
Therefore:
\begin{align*}
\mathbb{Z}/6\mathbb{Z}&=\{a+6\mathbb{Z} \mid a \in \mathbb{Z}\}\\
&=\{\dots,-3+6\mathbb{Z},-2+6\mathbb{Z},-1+6\mathbb{Z},6\mathbb{Z},1+6\mathbb{Z},2+6\mathbb{Z},\dots\}\\
&=\{\dots,-3-12,-3-6,-3,-3+6,-3+12,\dots,2-6,2,2+6,2+12,\dots\}\\
&=\{\dots,-15,-9,-3,3,9,\dots,-4,2,8,14,\dots\}
\end{align*}
However, the set $\mathbb{Z}/6\mathbb{Z}$ should be equal to the set $\mathbb{Z}_6=\{0,1,2,3,4,5\}$, which obviously isn't. Where am I doing mistake?
 A: $\mathbb{Z}/6\mathbb{Z} = \{\bar0,\bar1,\bar2,\ldots,\bar 5\} = \{6\mathbb{Z},\; 1+6\mathbb{Z},\; 2+6\mathbb{Z},\;\ldots,\;5+6\mathbb{Z}\}$
or if you prefer
$=\{\{\ldots,-12,-6,0,6,12,18,\ldots\},\{\ldots,-11,-5,1,7,13,\ldots\},\ldots,\{\ldots,-1,5,11,\ldots\}\}$
You cannot just collapse the layers of brackets.  $\mathbb{Z}/6\mathbb{Z}$ is a set with six elements; each of these elements is a set with an infinite number of elements (one equivalence class for the congruence-mod-6 equivalence relation).  In your question, the equal sign with a "$?$" on top is wrong, and in the last chain of equalities, the third (penultimate) one is wrong.
Of course, $\mathbb{Z}/6\mathbb{Z}$ also contains $\bar 6 = 6+6\mathbb{Z}$, but this is the same as $\bar 0 = 6\mathbb{Z}$, and similarly $\bar 7=\bar 1$ and so on (that's the reason for taking the quotient in the first place).  There are just six different elements.
A: Remember, whenever you divide you are forming "equivalence relations". That is: if I had 12 squares of chocolate to be divided between 3 people, I would number them and assign the 1st, 4th, 7th, 10th to Pete's box; the 2nd, 5th, 8th, 11th to Bob's; and the 3rd, 6th, 9th, 12th to Bill's. 
I am declaring two squares "equivalent" (belonging to the same person) if their difference is a multiple of three, thus I end up with 3 "equivalence classes" (the resulting boxes).
Now, if I had infinitely many balls to be divided between 6 people, again I would form equivalence classes. I would number them and then colour the ..., -11th, -5th, 1st, 7th, 13th,... blue; the ...,-10th, -4th, 2nd, 8th, 14th, red; etc. so that the infinite number of balls come in 6 colours, each assigned to one person. I am declaring two balls "equivalent" if their difference is a multiple of 6.
$a \sim b \hskip5pt iff \hskip5pt a-b \in 6\mathbb{Z}$
As a result, I end up with 6 equivalence classes, each consisting of infinitely manny balls. So you see how $\mathbb{Z}/6\mathbb{Z}$ has only 6 elements, where each of these elements is an infinite set (equivalence class), consisting of all the numbers whose difference lies in $6\mathbb{Z}$.
