# Is there a bijection between normal subgroups and quotient groups?

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.

• If you see the elements of a quotient $G/N$ as the cosets $Ng$ for $g \in G$, then $N$ is exactly the identity element of $G/N$. This gives you an inverse of the application $f$. Commented Sep 10, 2020 at 18:45
• If $N_1g_1=N_2g_2$, then I believe that $N_1$ and $N_2$ must be equal. What if you relax the condition of equality of quotient groups to one of Isomorphism of quotient groups ? Commented Sep 10, 2020 at 18:46
• @Simon Then you should probably also relax the condition of equality of normal subgroups to one of Isomorphism of groups ! Commented Sep 10, 2020 at 18:50

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 $$N_1=N_2$$.
• I see, because cosets partition the group, so two cosets are either equal or disjoint... so if two cosets intersect (in this case, at $e$), then they must be equal. Commented Sep 10, 2020 at 19:21
• All of that is true, but in fact we only need a simpler set-theory argument: Suppose $Q_1$ and $Q_2$ are sets of sets, and suppose $x$ is an object such that there exists a unique $S_1\in Q_1$ such that $x\in S_1$ and a unique $S_2\in Q_2$ such that $x\in S_2$. Then if $Q_1=Q_2$, we must have $S_1=S_2$. Commented Sep 10, 2020 at 19:45
• I'm curious if this statement could be modified to show $N_1\cong N_2$ (i.e. group isomorphic) under the precondition of $G/N_1\cong G/N_2$? Or even stronger, under the precondition $G_1/N_1\cong G_2/N_2$ given $G_1\cong G_2$? Or would this not be true? Commented May 20 at 1:05
• The "stronger" precondition is really the same, because one can use the isomorphism to work entirely within $G_1$ say. Commented May 20 at 1:40