03-03. The isomporphism theorem

Theorem 18. the second or diamond isomorphism theorem

• $G$,
  • group
• $A, B \subseteq G$,
  • subgroups
• $A \subseteq N_{G}(B)$,
  • subgroup

Then $A \cap B$ is sub group of $G$ and

• (0) $AB$ is subgroup of $G$ and
• (1) $B \trianglelefteq AB$,
• (2) $A \cap B \trianglelefteq A$,
• (3) $AB/B \cong A/(A\cap B)$,

proof.

(0) By Corollary 15, $AB$ is subgroup of $G$.

(1) It’s easy to see that

\[\forall b \in B, \ bBb^{-1} \subseteq B \Rightarrow B \subseteq N_{G}(B) .\]

Since $A \trianglelefteq N_{G}(B)$ by assumption and above, it follows that

\[\begin{eqnarray} \forall ab \in AB, \ abBb^{-1}a^{-1} & = & aBa^{-1} \quad (\because B \subseteq N_{G}(B)) \nonumber \\ & = & B \quad (\because A \subseteq N_{G}(B)) \nonumber \end{eqnarray} .\]

Hence $AB \subseteq N_{G}(B)$, that is, $B \trianglelefteq AB$.

(2)

\[\begin{eqnarray} \forall a \in A, \ a(A \cap B)a^{-1} & = & (A \cap B) \quad (\because A \cap B \subseteq B,\ A \subseteq N_{G}(B)) \nonumber \end{eqnarray} .\]

(3)

Since (1), quotient group $AB/B$ is well-defined. Define the map

\[\begin{eqnarray} \phi: A \rightarrow AB/B, \ \phi(a) := aB \nonumber \end{eqnarray} .\]

$\phi$ is homomorphism. Define natural projection $\pi$

\[\pi: AB \rightarrow AB/B, \ \pi(ab) := abB .\]

It follows that

\[\left.\pi\right|_{A} = \phi .\]

Indeed, let $a \in A$.

\[\begin{eqnarray} A \subseteq AB & \Rightarrow & \{z \in AB \mid za^{-1} \in B\} \supseteq \{z \in A \mid za^{-1} \in B\} \nonumber \\ \forall xy \in AB, \ x \in A, \ y \in B, \ & \Rightarrow & xya^{-1} \{z \in AB \mid za^{-1} \in B\} \supseteq \{z \in A \mid za^{-1} \in B\} \nonumber \end{eqnarray}\]

(4)

$\Box$