The correspondence theorem in my notes is:
Let $N < G$, $\pi = \text{can}:G \rightarrow G/H$. The map $H \rightarrow \pi(H)$ is a bijection between subgroups of $G$ containing $N$ and subgroups of $G/N$. Under this bijection, normal subgroups match with normal subgroups.
I've been asked to prove this theorem using the following three facts:
- $N \triangleleft G$ and $K \leq G/N$ then $\pi^{-1}(K) \triangleleft G \iff K \triangleleft G/N$
- $N \triangleleft G$ and $N \leq H \leq G \implies H = \pi^{-1}\pi(H)$
- Since $\pi$ is surjective, $K \subseteq G/N \implies K = \pi\pi^{-1}(K)$
Firstly, I want to prove that the map $H \rightarrow \pi(H)$ is bijective. Using $2$, I have that $H = \pi^{-1}\pi(H)$ which means that $\pi^{-1}$ is surjective when its pre-image is restricted to $\pi(H)$. in particular this means that the map $\pi(H) \rightarrow H$ is surjective, which is equivalent to $H \rightarrow \pi(H)$ being injective.
As for surjection, the map $H \rightarrow \pi(H)$ is clearly surjective by the construction of the map.
Now using $1$, if $K \triangleleft G/N \iff \pi^{-1}(K) \triangleleft G$. Therefore Normal subgroups "match".
I haven't used $3$, so I've definitely done something wrong. I would appreciate some help on where I'm going wrong.
Best Answer
I think I found the answer, so I'll post it here.
Firstly I'd like to prove that the map $\phi :H \rightarrow\pi(H)$ is bijective, where $\pi$ is the map $G \rightarrow G/N$. This $\phi$ bijectively maps subgroups of $G$ containing $N$ to subgroups of $G/N$.
Injective: $\phi(H_1) = \phi(H_2) \implies \pi(H_1) = \pi(H_2) \implies \pi^{-1}\pi(H_1) = \pi^{-1}\pi(H_2)$. Now by applying $2$ to each side of this equality, I have that $H_1 = H_2$.
Surjective: Take $A/N \leq G/N$. By $3$ I have that $\pi\pi^{-1}(A/N) = A/N$.
Furthermore, $A/N \leq G/N \implies N \triangleleft A \leq G$. Therefore $\pi^{-1}(A/N) = A$ which is a subgroup of $G$ containing $N$. so $\phi(A) = \pi(A) = A/N$. Therefore the map is also surjective.
Finally, I want to show that normal subgroups in this map "match" with each other. This immediately follows from $1$.