📂Abstract Algebra

Proof of the Third Isomorphism Theorem

Theorem ¹

Let $G,G'$ be a group.

• First Isomorphism Theorem: If there exists a homomorphism $\phi : G \to G'$ then $$ G / \ker ( \phi ) \simeq \phi (G) $$
• Second Isomorphism Theorem: If $H \le G$ and $N \triangleleft G$ then $$ (HN) / N \simeq H / (H \cap N) $$
• Third Isomorphism Theorem: If $H , K \triangleleft G$ and $K \leq H$ then $$ G/H \simeq (G/K) / (H/K) $$

The Isomorphism Theorem refers to these three independent theorems proven by the algebraist Emmy Noether.

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• $\ker$ is the kernel.
• $N \triangleleft G$ implies that $N$ is a normal subgroup of $G$.

Description

The Isomorphism Theorem, proven by the algebraist Emmy Noether, refers to these three independent theorems. Modifying the representation of the quotient group in the Third Isomorphism Theorem slightly gives $$ {{G} \over {H}} \simeq {{ {{G} \over {K}} } \over { {{H} \over {K}} }} $$ This is similar to multiplying both numerator and denominator by $\displaystyle {{1} \over {K}}$.

Proof

The proof of the Second Isomorphism Theorem is almost identical, differing only in the definition of $\phi$ and showing that $H$ is the kernel. Let’s define $\phi : G \to (G/K) / (H/K)$ as $\phi (g) = gK (H / K)$.

After showing that $\phi$ is a canonical mapping and that $H$ is $\ker \phi$, using the First Isomorphism Theorem completes the proof.

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Part 1. $\phi$ is a function.

For $x,y \in G$ $$ \begin{align*} && x= y \\ \implies& x K (H / K) = y K (H / K) \\ \implies& \phi (x) = \phi (y) \end{align*} $$ thus, $\phi$ is a function.

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Part 2. $\phi$ is a homomorphism.

For $x,y \in G$ $$ \begin{align*} \phi ( xy ) =& (xyK) ( H / K) \\ =& [ ( a K ) ( b K) ] ( H / K) \\ =& [ (aK) (H / K)] [ (bK) (H / K) ] \\ =& \phi ( x ) \phi ( y ) \end{align*} $$ thus, $\phi$ is a homomorphism.

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Part 3. $\phi$ is surjective.

For every $gK (H / N) \in (G / K) / (H / K )$ $$ \phi ( g ) = g K ( H / K ) $$ there exists some $g \in G$ satisfying this, so $\phi$ is surjective.

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Part 4. $H = \ker ( \phi )$

If $( \subset )$ $h \in H$ then $\phi (h) = h K ( H / K ) = K ( H / K )$ hence $$ h \in \ker ( \phi ) $$

If $( \supset )$ $h \in \ker ( \phi)$ then from $\phi (h) = K ( H / K ) = hK (H / K )$ $hK = K \in ( H / K )$ hence $$ h \in H $$

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Part 5.

First Isomorphism Theorem: If homomorphism $\phi : G \to G'$ exists, then $G / \ker ( \phi ) \simeq \phi (G)$

Since $\phi : G \to (G/K) / (H/K)$ is a homomorphism and surjective $$ \phi ( G ) = (G/K) / (H/K) $$ Meanwhile, because of $H = \ker ( \phi )$, by the First Isomorphism Theorem, the following holds. $$ G/H \simeq (G/K) / (H/K) $$

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