Normal correspondence theorem

Dependencies:

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

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

Proof of 1

Let $gN \in G/N$ and $kN \in K/N$.

\begin{align} & (gN)(kN)(gN)^{-1} \\ &= (gN)(kN)(g^{-1}N) \\ &= (gkg^{-1})N \\ &= k'N \tag{where $k' \in K$, because $K$ is normal in $G$} \\ &\in K/N \end{align}

Since $\forall gN \in G/N, \forall kN \in K/N, (gN)(kN)(gN)^{-1} \in K/N$, $K/N$ is normal in $G/N$.

Proof of 2

Let $K/N$ be a normal subgroup of $G/N$. We must prove that $K$ is normal in $G$.

Let $g \in G$ and $k \in K$.

\begin{align} & (gN)(kN)(gN)^{-1} \in K/N \tag{$K/N$ is normal in $G/N$} \\ &\Rightarrow (gkg^{-1})N \in K/N \\ &\Rightarrow gkg^{-1} \in K \end{align}

Since $\forall g \in G, \forall k \in K, gkg^{-1} \in K$, $K$ is normal in $G$.

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:

Third isomorphism theorem

Info:

Depth: 6
Number of transitive dependencies: 12