How is "integers ${\rm mod} ~n$ under addition" a cyclic group? I'm new to abstract algebra and group theories. I'm trying to study cyclic groups. For example, i know that integers $\mathbb Z+$ is a cyclic group since it is generated by $1$ (its additive identity, additive inverse and multiples). 
But if we take integers mod $n$ under addition, It will be a finite cyclic group:
$\{0,1, 2, 3, 4, 5,\ldots,n-1\}$
Now I know that integers $\pmod{n}$ under addition is a cyclic group since it is also generated by $1$. I also know that:
$n\equiv 0\pmod{n}$
$n+1\equiv 1\pmod{n}$
$n-1\equiv -1\pmod{n}$
But as I know, somehow, group generated by $1$ repeats itself and cycles through certain numbers:
$\ldots-2,  -1,  0,  1,  2,\ldots , n-1 , n , n+1 , n+2,\ldots, 2n-1, 2n, 2n+1, \ldots$
How does this group cycle? At which number does the group repeat itself?
Thank you!
 A: A cyclic group is defined as a group which can be generated by a single element.
For example, let $G = \Bbb Z/5\Bbb Z = \lbrace 0, 1, 2, 3, 4\rbrace$ with addition. Then
$$\begin{align}
1 &\equiv 1 \bmod 5\\
1 + 1 &\equiv 2 \bmod 5\\
1+1+1 &\equiv 3 \bmod 5\\
1+1+1+1 &\equiv 4 \bmod 5\\
1+1+1+1+1 &\equiv 0 \bmod 5
\end{align}$$
and since these are all of the elements of $G$, we conclude that this group is cyclic, and is generated by $1$, which can be written $G = \langle 1\rangle$ (although this requires some context, simply writing $\langle 1\rangle$ doesn't give enough information).
The way the elements cycle comes from the way the group $\Bbb Z/n\Bbb Z$ is defined. The "number" in the set $\lbrace 0, 1, 2, \dots, n-1\rbrace$ actually refers to the remainder when an integer is divided by $n$. For example, in the case $n = 5$, we see that $23 = 4 \times 5 + 3$, so $23 \equiv 3 \bmod 5$. Since the only possible candidates for remainders after dividing by $5$ are the numbers $0, 1, 2, 3, 4$ (convince yourself of this; for example, if I wrote $23 = 3 \times 5 + 8$, I could take another $5$ from the $8$ and have $23 = 3 \times 5 + 5+ 3 = 4\times 5 + 3$), these are the elements of the group.
A: Well, you said yourself that $0$ and $n$ are the same mod $n$.  So the group "cycles" when you reach $n$, since that's the same as going back to $0$.  In other words, if you list the elements you get by repeatedly adding $1$, they go: $$0,  1,  2,  \dots  , n-1 , 0 , 1 , 2, \dots , n-1, 0, 1, \dots$$
