The Third Isomorphism Theorem (The Double Fraction Theorem)

Theorem

Suppose $H, K ⊴ G$ (see Normal Subgroup) and $H \leq K$ . Then:

Since $\frac{K}{H} = {k H | k \in K}$ (see Quotient Group) then using $π_{H} : G \to \frac{G}{H}$ such that $g \mapsto g H$ has $π_{H} (K) = \frac{K}{H}$ and then $\frac{K}{H} ⊴ \frac{G}{H}$ .
$\frac{(\frac{G}{H})}{\frac{K}{H}} ≅ \frac{G}{K}$

Example

Like for the other theorems consider $\frac{Z}{24 Z}$ .

Here $H = ⟨ 8 ⟩$ and $K = ⟨ 4 ⟩$ . Here:

(\frac{\frac{(\frac{Z}{24 Z})}{H}}{\frac{K}{H}}) = {(0 + H) + \frac{K}{H}, (1 + H) + \frac{K}{H}, \dots, (3 + H) + \frac{K}{H}}

Which equals:

\frac{(\frac{Z}{24 Z})}{K} = the same thing

Proof

Proof

The strategy is to find a surjection $φ : \frac{G}{H} \to \frac{G}{K}$ with $Ker (φ) = \frac{K}{H}$ defined by $g H \mapsto g K$ . If you're worried about well-definedness, it turns out this will just work.

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