Let $G$ be a group. Is there a bijection from the collection of all normal subgroups of $G$, $\{ N: N \trianglelefteq G \}$, to the collection of all quotient groups of $G$ by normal subgroups, $\{ G/N: N \trianglelefteq G \}$?
My attempt: I tried to consider the obvious map $f$ that sends $N$ to $G/N$. Then $f$ is clearly surjective. However, I don't know whether it is injective. If $f(N_1) = G/N_1 = G/N_2 = f(N_2)$, we want to show that $N_1 = N_2$. I tried to show the contrapositive, i.e. let's suppose that $N_1 \neq N_2$. WLOG suppose that there is $g_1 \in N_1$ but $g_1 \notin N_2$. Since $G/N_1 = G/N_2$, we know that $N_1g_1 = N_2g_2$ for some $g_2$. This is where I am stuck. I don't know if that tells me anything.
The elements of $G/N_1$ are the cosets of $N_1$. Exactly one of those, namely $N_1$ itself, contains the identity element $e$ of $G$. Similarly, exactly one element of $G/N_2$, namely $N_2$ itself, contains $e$. Therefore if $G/N_1 = G/N_2$, then we would be forced to have $N_1=N_2$.