Suppose $G$ is a finite group and $N$ is a normal subgroup of $G$. Then subgroups of $G/N$ are of the form $A/G$ for $A\le G$. But how does the normal subgroups of $G/N$ look like? Is it true that $A \unlhd G \iff A/N \unlhd G/N$ provided $N$ is a subgroup of $A$?
I can see that if $A\unlhd G$ and $N\unlhd A$ then $A/N \unlhd G/N$.
If I then assume that $A/N \unlhd G/N$ and take $xN \in A/N$ and $yN \in G/N$ then $(xN)^{(yN)} = (x^y)N \in A/N$. Does this imply that $x^y \in A$?
You're definitely on the right track. You need to show that $xN \in A/N \implies x \in A$.
Some hints to think about:
$1)$ What does it mean for $xN$ to be in $A/N$?
$2)$What does it mean for two cosets to be the same?
Hopefully this should help you out.