Group theory confusion for $|\langle g\rangle| = o(g)$ From this intro to group theory text I'm reading:
The subgroup generated by $g$ is $\{g^n\ |\ n\in G \ and \  n \in \mathbb{Z}\}$
So if we take $\mathbb{Z}_6$ = $(\{0, 1, 2, 3, 4, 5\}, +)$
And the group generated by the element $ \langle 2\rangle = (\{0, 2, 4\}, +)$
Then $|\langle 2\rangle | = 3$
But then the text gives the equation $|\langle g\rangle | = o(g)$
$|\langle2 \rangle| = o(g)$ = the smallest $n$ where $2n = e$ ($e$ is $0$ for $\mathbb{Z}$)
Here is where I get confused.
If $|\langle 2 \rangle|$ is $3$, then $o(2)$ must be $3$ as well, but $2^0 = e$.
So the smallest $n$ such that $2^n = e = 0$?
Where am I going wrong?
 A: They ought to have specified: the order of an element $g$ of a group $G$ is the smallest positive number $n$ such that $g^n = e$. This is so that the statement is not vacuous; in a group every element $g$ satisfies $g^0 = e$.
A: Here $\langle 2\rangle = \{2,4,0\}$ so $|\langle 2\rangle| = 3$.  Also $o(2) = 3$;  that is $3$ is the smallest positive integer such that $2\times 3 = 0$ in this group.  We do not consider $n=0$ when we look for the smallest $n$ so that $2 \times n = 0$.
A: You are a bit confused because in the example you gave the group is in additive notation and in text they use multiplicative notation.
Try to think about the operation as $*$. The order of an element $g$ written $o(g)$ is the answer you get when you apply $*$ on $g$ repeatedly unitl you get $e$.
That is in $\mathbb{Z}_6$ $o(2)=3$ since this is an additive group we think about the operation as addition and get
$$
2+2+2=6\equiv_6 6.
$$
To give an example of a multiplicative group take $\mathbb(Z)_5^*$ which consists of the numbers $\{1,2,3,4 \}$ and we use multiplication as the operation and as $e$ we have $1$. Here $o(2)=4$ since
$$
2^4=2\cdot 2\cdot 2 \cdot 2 \cdot 2= 16\equiv_5 1
$$
