The Second Isomorphism Theorem (The Diamond Iso. Theorem)

Theorem

Suppose $A, B \leq G$ and $A \leq N_{G} (B)$ (see Normalizer, so every element of $A$ normalizes $B$ , so $a B a^{- 1} = B$ for all $a \in A$ ). Then:

$A B = ⟨ A \cup B ⟩$ is the smallest subgroup containing $A$ and $B$ . It is the "tip" of the diamond in the lattice.
$A \cap B ⊴ A$ and $B ⊴ A B$
$\frac{A B}{B} ≅ \frac{A}{A \cap B}$

This may be the most useless theorem ever, but the last finding is really important. Namely, for when you want to "cancel" $B$ from $A B$ then you have to consider the factor of $A \cap B$ (ie: you get some elements of $B$ that "bleed over")

Note

If $B ⊴ G$ (see Normal Subgroup) then $N_{G} (B) = G$ so $A \leq N_{G} (B)$ .

Example

Here $A = ⟨ 6 ⟩$ and $B = ⟨ 4 ⟩$ . Then $A B = ⟨ 2 ⟩$ and $A \cap B = ⟨ 12 ⟩$ .

If you take $A B$ and divide it into the cosets, you can make pairings between cosets of $A B$ versus that of just $A$ w.r.t $A \cap B$ .

Proof

Proof

(Sketch of (3)): Define a map $φ : A \to \frac{A B}{B}$ such that $a \mapsto a B$ (note that $a B \in \frac{A B}{B}$ still, our notation still makes sense). It's a Homomorphism (or Group Morphism) since $a_{1} B \cdot a_{2} B = a_{1} a_{2} B$ so $φ (a_{1} a_{2}) = φ (a_{1}) φ (a_{2})$ .

Note that $φ$ is surjective since if you take any $(a b) B$ for some $a b \in A B$ , then check that:

(a b) B = a B

since $a^{- 1} (a b) = b \in B$ via Left (and Right) Cosets form a Partition in G. Then $φ (a) = a B = (a b) B$ so then $φ$ is surjective.

Notice also $Ker (φ) = {a \in A | a B = 1 B}$ . But $a B = 1 B ⟺ a \in B$ , so then:

Ker (φ) = {a \in A | a \in B} = A \cap B

showing injectivity.

Using the The First Isomorphism Theorem of Groups, The Fundamental Theorem for Group Morphisms, then:

\frac{A}{Ker (φ)} = \frac{A}{A \cap B} ≅ \frac{A B}{B}

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