Understanding why the order of $3+\langle 6\rangle$ in $\mathbb{Z}_{15}/\langle 6\rangle$ is $1$ I'm trying to understand the previous thread (link) which finds the order of $3+\langle 6\rangle$ in $\mathbb{Z}_{15}/\langle 6\rangle$.
I was following the logic from that thread. First question is why in a group $(\mathbb{Z}_{15},+)$ we get:
$$(3+\langle6\rangle)^{n}=\underset{n\,times}{\underbrace{(3+\langle6\rangle)+(3+\langle6\rangle)+...+(3+\langle6\rangle)}}=3n+\langle6\rangle$$
An not $3n+n\langle 6\rangle$  (if we do get it and then why $n\langle 6\rangle =\langle 6\rangle$)?
Second question is about the line: 

Now note that the identity of $\mathbb{Z}_{15}/\langle 6\rangle$ is $0 + \langle 6\rangle = \langle 6\rangle$, so the order of $3 + \langle 6\rangle$ is the smallest positive integer $3n + \langle 6\rangle = \langle 6\rangle$.

What is the identity of $\mathbb{Z}_{15}/\langle 6\rangle$? Is it $0$ or $\langle 6\rangle$? I feel like the author wanted to point out that $\langle 6\rangle$ is the identity element but why $0$ is not? Also later it said that $3+\langle 6\rangle$ is the identity element. How is it possible to have more then one identity element in a group?
Last question is why $\langle 6\rangle=\{0,3,6,9,12\}$? From this thread I learn that if order of a group is $15$ then total number of generators of group $G$ are equal to positive integers less than $15$ and co-prime to $15$. So for example $3$ and $9$ are not in $\langle 6\rangle$.
I feel confused and it feels like I miss the definition of the cyclic group (even if it in front of me). Will be glad for some clarifications. Also, sorry for wasting time on explaining basic things.
 A: The addition in a quotient group $\mathbb{Z}_{15}/\langle 6\rangle$ is defined as
$$(a+\langle 6\rangle) + (b + \langle 6\rangle) = (a+b) + \langle 6 \rangle$$
for $a,b \in \mathbb{Z}_{15}$. Hence no adding of $\langle 6\rangle$s.
On the other hand, equality in the quotient group is defined as $a + \langle 6\rangle = b + \langle 6\rangle$ if $a-b \in \langle 6\rangle$. The identity element of $\mathbb{Z}_{15}/\langle 6\rangle$ is $0 + \langle 6\rangle$, which is commonly denoted simply as $\langle 6\rangle$.
Let's determine $\langle 6\rangle$. Certainly $0,6 \in \langle 6\rangle$. Since a subgroup is closed under inverting we also have $-6 \in \langle 6\rangle$. But we have $-6 = 9$ in $\mathbb{Z}_{15}$ (the equality actually being congruence mod $15$) so $9 \in \langle 6\rangle$.
Since a subgroup is closed under addition, we have $12 = 6 + 6 \in \langle 6\rangle$. And then again $3 = -12 \in \langle 6\rangle$.
So far we got $\{0,3,6,9,12\} \subseteq \langle 6\rangle$. Now check that $\{0,3,6,9,12\}$ is indeed a subgroup of $\mathbb{Z}_{15}$, i.e. that the inverse of every element and the product of every two elements from the set stays in the set (working mod $15$ of course). Since $\langle 6\rangle$ is the smallest subgroup of $\mathbb{Z}_{15}$ containing $6$, we conclude $\langle 6\rangle \subseteq \{0,3,6,9,12\}$, and hence they are equal.
Finally, we see that $3 + \langle 6\rangle = 0 + \langle 6\rangle$ because $3-0 = 3 \in \langle 6\rangle$. Therefore $3 + \langle 6\rangle$ is indeed the identity element $\langle 6\rangle$ in $\mathbb{Z}_{15}/\langle 6\rangle$. 
