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.

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 $3a=3\neq 0$...
