Correspondence theorem

Dependencies:

Let $N$ be a normal subgroup of $G$. Then:

If $K$ is a subgroup of $G$, then $K/N$ is a subgroup of $G/N$.
Every subgroup of $G/N$ is of the form $K/N$, where $K$ is a subgroup of $G$.
Let $\phi(K) = K/N$. Then $\phi$ is a bijection from the subgroups of $G$ containing $N$ to subgroups of $G/N$.

Proof of 1

Let $k_1N, k_2N \in K/N$.

\begin{align} & (k_1N)(k_2N)^{-1} \\ &= (k_1N)(k_2^{-1}N) \\ &= (k_1k_2^{-1})N \\ &\in K/N \end{align}

Therefore, $K/N$ is a subgroup of $G/N$.

Proof of 2

Let $H$ be a subgroup of $G/N$. Let $K = \{k: kN \in H\}$.

\begin{align} & k_1, k_2 \in K \\ &\Rightarrow k_1N, k_2N \in H \\ &\Rightarrow (k_1N)(k_2N)^{-1} \in H \\ &\Rightarrow (k_1k_2^{-1})N \in H \\ &\Rightarrow k_1k_2^{-1} \in K \end{align}

Therefore, $K$ is a subgroup of $G$ and $H = K/N$.

Proof of 3

By result 2, $\phi$ is onto.

Let $\phi(K_1) = \phi(K_2) \Rightarrow K_1/N = K_2/N$.

\[ k_1 \in K_1 \Rightarrow k_1N \in K_1/N = K_2/N \Rightarrow k_1 \in K_2 \Rightarrow K_1 \subseteq K_2 \]

Similarly, $K_2 \subseteq K_1$. Therefore, $K_1 = K_2$, which means $\phi$ is one-to-one.

Therefore, $\phi$ is a bijection.

Dependency for:

Normal correspondence theorem

Info:

Depth: 5
Number of transitive dependencies: 11