Different way's of approaching the order of an element When I was making one of the assignments from the book 'Groups and Symmetry' by M. A. Armstrong, I got a little confused about the order of an element of a group.
First there was exercise 4.2, where you had to find the order of each element of the group $\mathbb{Z}_{12}$, this got me the following:
elements $1, 5, 7, 11$ have order $12$
elements $2, 10$ have order $6$
elements $3, 9$ have order $4$
elements $4, 8$ have order $3$
element $6$ has order $2$  
But then in the next assignment you're dealing with the following group of integers {$1, 2, 4, 7, 8, 11, 13, 14$} under multiplication modulo 15, where the order of each element is:
element $1$ has order $1$
elements $2, 7, 8, 13$ have order $4$
elements $4, 11, 14$ have order $2$
Now my question is, if we look at for example element $2$ from each group:
$2\cdot 6=12=0(mod12)$, this gives the the element $2$ from the first group order $6$
$2^{ 4 }=16=1(mod15)$, this gives the element $2$ from the second group order $4$
How can it be that in the first case we're getting the order of the element when we end up with $0$, but in the second case we end up with $1$?
Is it because $0$ and $1$ are the smallest possible elements in the group and that's what we have to 'work towards to'?   
Or do I interpret this in a wrong way and should I approach it in a different way (maybe because the second group is under multiplication modulo n)?
 A: In the first group, the identity element is $0$ where as in the second group the identity element is $1$. The order of an element $g \in G$ is the smallest $n \in \mathbb{N}$ such that $g^n=e$ where $e$ is the identity element of $G$. 
Note: In Example-1, $e=0$ and the operation is addition. So $g^n$ turned out to be $ng$. In Example-2, $e=1$ and the operation is multiplication modulo $15$.  
A: The order is defined based on the identity element. For a group $G$ with operation $\oplus$ and identity element $id_G$ the order of $g$ is given by the smallest positive $k$ such that
$$\underbrace{g\oplus \cdots\oplus g}_{k-times}=id_G$$
In the first example, the group operation is addition so $0$ is the identity. In the second it is multiplication so $1$ is the identity.
A: There are two issues. The first one is, that the neutral element in a group written additively is $0$, and multiplicatively written is $1$. So $g^n=1$ corresponds to $ng=0$. Secondly, from $g^n=1$, or $ng=0$ it does not follow, that $g$ has order $n$. Only, that the order is a divisor of $n$.
