In this problem we work out step-by-step a complete set of left coset representatives of the subgroup $9 \mathbb{Z}$ of $\mathbb{Z}$. First, let's recall the definition of coset representatives.

Let $H$ be a subgroup of a group $G$. Then a list of elements

$g_1, g_2,...$

is called a complete set of left-coset representatives of $H$ in $G$ if
a) for every element $g$ in $H$ there exists some $g_i$ such that $gH = g_i H$
b) if $i \neq j$ then $g_i H \neq g_j H$.

Now, take $G = \mathbb{Z}$ and $H = 9 \mathbb{Z}$. Since $G$ is an additive group, for any two elements $x, y$ in $G$, the coset of $x$ and the coset of $y$ are the same if and only if

x+y is in H
xy is in H
x-y is in H
x/y is in H

It follows that every element of $\mathbb{Z}$ is represented by an integer $z$ in the range (NOTE: there could be more than one correct answers; select all the correct ones):

A. $-5 < z < 5$

B. $0 \leq z < 9$

C. $3 < z < 6$

D. $2 \leq z \leq 11$

Finally, write down a complete set of left representatives of $9 \mathbb{Z}$ in $\mathbb{Z}$. Make sure that each representative is a non-negative integers $< 9$.
Enter your answer as a comma separated list.