Second isomorphism theorem

Dependencies:

Let $G$ be a group. Let $S$ be a subgroup of $G$ and $N$ be a normal subgroup of $G$. Then the following hold:

$SN$ is a subgroup of $G$.
$S \cap N$ is a normal subgroup of $S$.
$SN/N \cong S/(S \cap N)$.

Proof or result 1

$SN$ is a non-empty subset of $G$.

Let $s_1n_1, s_2n_2 \in SN$ where $s_1, s_2 \in S$ and $n_1, n_2 \in N$.

\begin{align} & (s_1n_1)(s_2n_2)^{-1} \\ &= s_1n_1n_2^{-1}s_2^{-1} \\ &= s_1ns_2^{-1} \tag{where $n = n_1n_2^{-1} \in N$} \\ &= s_1s_2^{-1}n' \tag{where $n' \in N$, because $N$ is normal in $G$} \\ &= sn' \tag{where $s = s_1s_2^{-1} \in S$} \\ &\in SN \end{align}

Therefore, $SN$ is a subgroup of $G$.

Proof of result 2

$S \cap N$ is a non-empty subset of $S$.

\begin{align} & h_1, h_2 \in S \cap N \\ &\Rightarrow h_1, h_2 \in S \wedge h_1, h_2 \in N \\ &\Rightarrow h_1h_2^{-1} \in S \wedge h_1h_2^{-1} \in N \tag{because $S$ and $N$ are subgroups of $G$} \\ &\Rightarrow h_1h_2^{-1} \in S \cap N \end{align}

Therefore, $S \cap N$ is a subgroup of $S$.

Let $s \in S$ and $h \in S \cap N$.

$shs^{-1} \in S$, because of closure of $S$. $shs^{-1} \in N$, because $N$ is normal in $G$.

Therefore, $shs^{-1} \in S \cap N$, which means $S \cap N$ is normal in $S$.

Proof of result 3

\[ SN/N = \{snN: s \in S, n \in N\} = \{sN: s \in S \} \]

Let $\psi: S \mapsto G/N$ such that $\psi(s) = sN$. Then $\psi(S) = \{sN: s \in S\} = SN/N$.

\begin{align} & \psi(s_1s_2) \\ &= (s_1s_2)N \\ &= (s_1N)(s_2N) \tag{because $N$ is normal in $G$} \\ &= \psi(s_1)\psi(s_2) \end{align}

Therefore, $\psi$ is a homomorphism.

\begin{align} & \psi(s) = \operatorname{id}(G/N) = N \\ &\Rightarrow sN = N \\ &\Rightarrow s \in N \\ &\Rightarrow s \in S \cap N \end{align}

Therefore, $\operatorname{ker}(\psi) = S \cap N$.

By the first isomorphism theorem, \[ \psi(S) \cong S/\operatorname{ker}(\psi) \Rightarrow SN/N \cong S/(S \cap N) \]

Dependency for: None

Info:

Depth: 7
Number of transitive dependencies: 17