If $N \triangleleft G$ and $a,b \in G$ then why does $aN=bN$ imply $a^{-1}b \in N$?
I don't find it too obvious, sadly, since $N=a^{-1}bN$ by rearrangment, and this tells me that $a^{-1}b$ is the identity? Is that why it must be in $N$? That's the only reason I can think of but not sure whether or not it's true.
Is there a more rigorous reasoning behind it?
$aN=bN$ tells you that every element on the right hand side is also an element on the left hand side (and vice versa). As $1\in N$, one particular element on the right is $b\cdot 1=b$. We conclude $b\in aN$, i.e., there exists $n\in N$ such that $b=an$. But then $a^{-1}b=n\in N$.