The order of elements in the quotient group Let $G$ be a group, $N$ a normal subgroup of $G$, $a \in G$, and let $k = o(a)$.
I don't understand why the order of an element in $G/N$ is not necessarily equal to the order of the "corresponding" element in $G$ (i.e, why it might be that $o(a) \neq o(aN)$).
My reasoning is this: We know $k$ to be the smallest positive integer such that $a^k = e$. 
Let $m = o(aN)\Rightarrow (aN)^m=eN \Rightarrow (a^m)N=eN \Rightarrow a^m=e$.
If $m < k$, it is a contradiction to $k$ being $a$'s order.
If $m > k$, than $k$ is really $a$'s order since $(aN)^k=(a^k)N=eN=N$.
I know I'm wrong, but I am not sure where.
 A: The best example to illustrate this is, I believe, the infinite cyclic group $\mathbb{Z}$. Here, every non-trivial element has infinite order. Now, consider a quotient group, for example $\mathbb{Z}/3\mathbb{Z}$. This quotient group is cyclic, of order three. Thus, every element has order three.
In your proof your issue is with the following line.

If $m>k$, then $k$ is really $a$'s order since $(aN)^k=(a^k)N=eN=N$.

This does not prove that $k$ is $a$'s order. In our exmaple, every element having order three means that $3a\in3\mathbb{Z}$ for all $a\in\mathbb{Z}$, not that $3a=0$. The element $3a$ can be any element of $3\mathbb{Z}$, not just the trivial element. Take $a=1$, then $3a=3\neq 0$...
A: Recall that the cosets $aN$ are equivalence classes. When two cosets are equal, say $sN=tN$, it means both the representatives $s$ and $t$ are in the same equivalence class. It does not mean that the two are equal (they are simply equivalent). 
In particular, $sN=tN$ just implies $st^{-1} \in N$. This is not the same as saying $s=t$.
A: Just to consider an example related to the infinite cyclic one, let $G = \langle a \rangle$ be cyclic of order $n$, and $k$ be any divisor of $n$.
Consider $N = \langle a^{k} \rangle$. Then $aN$ has order $k$ in $G/N$.
A: Let $a\in G$ have finite order; it can be computed as the minimum $k>0$ such that $a^k=1$.
But the same holds for an element $aN\in G/N$: the order is the minimum exponent $h>0$ such that $(aN)^h=1N$, which means the minimum $h>0$ such that $a^h\in N$, because $(aN)^h=a^hN$ and $gN=1N$ if and only if $g\in N$.
Thus $a^k=1$ certainly implies that $a^k\in N$, but the converse may not be true.
If you use a different characterization of the order, namely the cardinality of the cyclic subgroup generated by the given element, you can be even more precise.
Consider the canonical projection $\pi\colon G\to G/N$. Then, certainly,
$$
\pi(\langle a\rangle)=\langle aN\rangle
$$
so, by the homomorphism theorems,
$$
\langle aN\rangle \cong \frac{\langle a\rangle N}{N}\cong 
\frac{\langle a\rangle}{\langle a\rangle \cap N}
$$
which implies that the order of $aN$ divides the order of $a$. The examples you find in other answers show that any divisor can result.
In the case when $a$ has infinite order, the above reasoning with the homomorphism theorem still holds, showing that the order of $aN$ can be any integer (or be infinite).
So what the order of $aN$ is strictly depends on both $a$ and $N$ and nothing more than “the order of $aN$ divides the order of $a$” can be said in general.
