I am using Fraleigh's text. Lemma 34.3 states,
Let $G$ be a group and let $N \unlhd G$. Let $\gamma : G \rightarrow G/N$ by the canonical hom. Then $\phi$ from the set of normal subgroups of $G$ containing $N$ to the set of normal subgroups of $G/N$ by $\phi(L) = \gamma[L]$ is a bijection.
The proof goes (revised for those who do not have the text handy, although you can find it online):
Since $L \unlhd G$ containing $N$, then $\phi(L) = \gamma [L] \unlhd G/N$.
Here's the line that's throwing me:
Because $N\leq L$, for each $x\in L$, the entire coset $xN$ in $G$ is contained in $L$. Thus, by Thm 13.15, $\gamma^{-1}[\phi(L)] = L$.
I need that ^ line explained.
I've provided the statement of Thm 13.15 below, as well as the rest of the 1-1 proof portion of the lemma.
Thm 13.15:
Let $\phi: G \rightarrow G^{'}$ be a group hom. Let $H=Ker(\phi)$. Let $a\in G$. Then,
$\phi^{-1}[\{\phi (a)\}] = \{x\in G | \phi(x) = \phi (a)\}$
is the left coset $aH$ of $H$ and is also the right coset $Ha$ of $H$.
Rest of 1-1 portion of proof:
Let $\phi (L_{1}) =\phi (L_{2})$ for $L_{1}, L_{2}\in Dom(\phi)$. Then since $\gamma^{-1}[\phi(L)] = L$ for any normal subgroup $L$ containing $N$, $L_{1}=L_{2}$.
By definition of the canonical map, for each $x\in G$, $\gamma(x)=xN$. Also $y\in \gamma^{-1}\bigl(\gamma(x)\bigr)$ means that $\gamma(y)=\gamma(x)$, i.e. $yN=xN$. In other words, $y\in xN$. Thus $\;\gamma^{-1}\bigl(\gamma(x)\bigr)=xN$, and finally $$\gamma^{-1}\bigl(\gamma(L)\bigr)=\bigcup_{x\in L}\gamma^{-1}\bigl(\gamma(x)\bigr)=\bigcup_{x\in L}xN\subset L\quad\text{if }N\subset L.$$ The reverse inclusion is obvious.