From Lee's Abstract Algebra, Theorem 4.8 (proof as exercise for the reader):
Let $G$ be a group and $N$ a normal subgroup. Then the subgroups of
$G/N$ are precisely of the form $H/N$, where $N \leq H \leq G$.
Note that $H \leq G$ means "$H$ is a subgroup of $G$."
This question begins exactly the same way as Showing that every subgroup of a factor group $G/N$ has the form $H/N$, but without using homomorphisms. To do this:
Let $J \subseteq G/N=\{aN : a \in G\}$, so $J=\{bN : b \in H \subseteq G \}$. So any subset $J$ of $G/N$ is of the form $J=H/N$ for $H \subseteq G$.
Now suppose that $J \leq G/N$, so $J$ is a group of the form $H/N$. Thus $eN \in J$, [$bN \in J \implies b^{-1}N \in J$], and [$b_1N,b_2N \in J \implies b_1 b_2 N=b_3N \in J$]. In other words, $e\in H$, $H$ contains inverses, and $H$ is closed under $G$'s binary operation. So any subgroup $J$ of $G/N$ is of the form $J=H/N$ for $H \leq G$.
Now how do I show that $N \subseteq H$ to complete the proof?
Here's some of my failed attempt: Let $n \in N$ and $H/N \leq G/N$, we must show that $n \in H$. Suppose that $n \notin H \leq G$, so $n^{-1} \notin H$. Then the cosets $nN=Nn$ and $n^{-1}N=Nn^{-1}$ are not in the group $H/N$ … stuck.
EDIT: I notice that $nN=N=eN$, so the set $nN$ is certainly in $H/N$, but I don't think this proves $n$ must be in $H$. Also, this seems to be part of the "correspondence theorem for groups," but I am having trouble understanding the proofwiki (https://proofwiki.org/wiki/Correspondence_Theorem_%28Group_Theory%29) and wonder if there's an easier way to show only the part that $N \subseteq H$.
Best Answer
The proofwiki proof goes as follows: let $J$ be a subgroup of $G/N$, so $J$ is some set of cosets of $N$. Define $\beta(J)=\{g \in G \mid gN \in J \} \subseteq G$; notice that one way of expressing $J$ is $J = \beta(J) / N$. And $J$ is a group, so $nN=N=eN \in J$, so $N \subseteq \beta(J) \subseteq G$. To complete the proof, show that $\beta(J)$ is a group, which I already did above.
I was confused earlier because I thought we had to prove that $J$ can only be expressed as $H/N$ for $N \leq H \leq G$. Maybe this stronger statement is also true; I asked about it here: Do alternative expressions exist for subgroups of $G/N$, distinct from the correspondence theorem for groups?. Regardless, the theorem is true: subgroups $J$ of $G/N$ are always expressible as $\beta(J)/N$ for $N \leq \beta(J) \leq G$. Perhaps there are other ways to express $J$ as $W/N$ for $W$ not containing $N$, but $J$ can always be written in the clean form of $H/N$ for $N \leq H \leq G$.