Every subgroup of a quotient group is a quotient group itself

Question

Let $G$ be a group and $N$ its normal subgroup. Now, let $B$ be a subgroup of $G/N$. I need to prove that $B = A/N$ for some subgroup $A$ of $G$ that contains $N$.

Here's what I did: Given a normal subgroup $N$, we have the canonical projection $\pi:G \to G/N$. Let $B$ be a subgroup of $G/N$. Then $A = \pi^{-1}(B)$ is a subgroup of $G$. $N \in B$, hence $\pi^{-1}(N) = N \subseteq A$. So, $N$ is a normal subgroup of $A$.

Not sure what to do next.

Answer 1

You're nearly there. Just use the set-theoretic fact that if $f : S \to T$ is a surjective function, and $B \subseteq T$, then $$ f( f^{-1}(B)) = B. $$

In your case, you obtain that $B = \pi(A) = A/N$.

Answer 2

You were almost done. Since $\pi$ is onto, you have $$A/N=\pi (A) = \pi (\pi^{-1}(B))=B$$