Q:

Cyclic groups generators, understanding example $\mathbb{Z}_6$

I am studying group theory but I am having hard time understanding the cyclic group $\mathbb{Z}_n$. On the material I am studying I have the following example:

$\mathbb{Z}_6=\{0_6,1_6,...,5_6\}$ and $1_6=7_6=13_6=-5_6=...=\{...,-11,-5,1,7,13,...\}$

I know that this equality $1_6=7_6=13_6=-5_6$ holds because each of the remainder is $1$. But I do not understand what ($\mathbb{Z}_6)=\{0_6,1_6,...,5_6\}$ means. 
Question:
1) Considering $\mathbb{Z}_6=\{0_6,1_6,...,5_6\}$. Is $\mathbb{Z}_6=\{0_6,1_6,...,5_6\}$ all the generators of $6$?
2) How can I generate $\mathbb{Z}_6$ from $5$? I am thinking each one of this numbers is a cycle to attain $\mathbb{Z}_6$. However if we consider $\langle 5 \rangle=5+5+5...+5$ as a cyclic subgroup of $(\mathbb{Z},+)$. How can $5_6$ generate $6$? How do I read $5_6$?
Thanks in advance!

A:

Let's assume your teacher and/or your text are using the (non-standard) notation $n_m$ to mean $n\pmod m$.  
The equation $\mathbb{Z}_6=\{0_6,1_6,...,5_6\}$ is an abuse of notation which is usually interpreted to mean that $\mathbb{Z}_6$ is a group with elements $\{0_6,1_6,...,5_6\}$ and group operation defined by $a_6 \star b_6 = (a+b)_6$. You can check that this is a group. It is not a subgroup of $\mathbb{Z}$ because its elements are not elements of $\mathbb{Z}$, its elements are in fact subsets of $\mathbb{Z}$. One of those subsets, $0_6$, is a subgroup of $\mathbb{Z}$. The others are not, they are called the "cosets of $0_6$ in $\mathbb{Z}$". But we don't  need to deal with them here. 
Your question 1: A set, $S$, of generators for a group, $G$, is a set with the property that every element of $G$ can be written a product of elements of $S$ and their inverses. Thus it looks like you may have a typo in this question since $6$ is not a group.  However, you can show that $6$ generates the subgroup $0_6$ of $\mathbb{Z}$. 
Your question 2: Observe that $5_6 \star 5_6 = 4_6$ and
$5_6 \star 4_6 = 3_6$ and
$5_6 \star 3_6 = 2_6$ and
$5_6 \star 2_6 = 1_6$ and
$5_6 \star 1_6 = 0_6$.
So every element of $\mathbb{Z}_6$ can be written as a product of the element of the single element set $\{5_6\}$ and thus it generates  $\mathbb{Z}_6$. By an abuse of language we say that the element $5_6$ generates $\mathbb{Z}_6$. Only one other element of  $\mathbb{Z}_6$ generates it. 

