Consider a group $G$, $N\trianglelefteq G$ and $H:=G/N$. Let $K\le H$. Then there exists a unique $A\le G$ that contains $N$, such that $K=A/N$. Moreover, $A=K^{\pi^{-1}}$, with $\pi:G\to H$ the canonical projection.
Proof:
Let $A\le G$ an arbitrary subgroup that contains $N$. Then $N\trianglelefteq A$ and $A$ is a union of cosets of $N$ (in $G$). Then, $A/N$ is well-defined and is exactly equal to this set of cosets. There is always at most one such subgroup $A$. ...
I don't know how to explain the bold sentence. Why is $A$ unique?
Thanks.
If $A/N = B/N$ with $A$ and $B$ both subgroups that contain $N$, then for any $a\in A$ there exists $b\in B$ such that $aN=bN$, hence $b^{-1}a\in N\subseteq B$, so $b(b^{-1}a)=a\in B$. Thus $A\subseteq B$. Symmetrically, $B\subseteq A$ so you get $A=B$.