📂Abstract Algebra

Proof of the Second Isomorphism Theorem

Theorem¹

Let $G,G'$ be groups.

• First Isomorphism Theorem: If a homomorphism $\phi : G \to G'$ exists, 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 the three independent theorems above, proven by the algebraist Emmy Noether.

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

Explanation

In most cases where the second isomorphism theorem is used, $HN := \left\{ hn \ | \ h \in H , n \in N \right\}$ may just as well be thought of as the product of two normal subgroups $H , N \triangleleft G$, in which case the fact that $HN \triangleleft G$ can be shown without difficulty through a simple computation. If we slightly rewrite the expression of the quotient group in the second isomorphism theorem, we get $$ {{HN} \over {N}} \simeq {{H } \over { H \cap N }} $$ This can be seen as a kind of ‘cancellation’, erasing the factor common to the numerator and the denominator.

Proof

Strategy: Define $\phi : HN \to H / ( H \cap N)$ as $$ \phi (hn) := h(H \cap N) $$ After showing that $\phi$ is a surjective homomorphism and that $N$ is $\ker \phi$, applying the first isomorphism theorem completes the proof.

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Part 1. $\phi$ is a function.

• Since the domain of $\phi$ is $HN$, if we let $x \in HN$, then it is expressed as the product $x = h_{x} n_{x}$ of some $h_{x} \in H$ and $n_{x} \in N$.
• Since the codomain of $\phi$ is $H / \left( H \cap N \right)$, the value $\phi (x)$ of $\phi$ at $x \in HN$ is expressed in the form $h_{x} \left( H \cap N \right)$, where some element $h_{x} \in H$ is multiplied on the left of the coset $\left( H \cap N \right)$ of $H$.

Therefore, for $x,y \in HN$, $$ \begin{align*} & x = y \\ \implies& h_{x} n_{x} = x = y = h_{y} n_{y} \\ \implies& H \ni h_{y}^{-1} h_{x} = n_{y} n_{x}^{-1} \in N \\ \implies& h_{y}^{-1} h_{x} \in \left( H \cap N \right) \\ \implies& h_{x} \in h_{y} \left( H \cap N \right) \\ \implies& h_{x} (H \cap N) = h_{y} (H \cap N) \\ \implies& \phi \left( h_{x} n_{x} \right) = \phi \left( h_{y} n_{y} \right) \\ \implies& \phi (x) = \phi (y) \end{align*} $$ so $\phi$ is a function.

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Part 2. $\phi$ is a homomorphism.

For $h_{1} n_{1} , h_{2} n_{2} \in HN$, $$ \begin{align*} \phi ( ( h_{1} n_{1} ) ( h_{2} n_{2} ) ) =& \phi ( h_{1} h_{2} n_{1} n_{2} ) \\ =& h_{1} h_{2} (H \cap N) \\ =& h_{1} (H \cap N) h_{2} (H \cap N) \\ =& \phi (h_{1} n_{1} ) \phi ( h_{2} n_{2} ) \end{align*} $$ Therefore, $\phi$ is a homomorphism.

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Part 3. $\phi$ is surjective.

Let $e$ be the identity element of $N$. Then for every $h(H \cap N) \in H / (H \cap N )$, there exists $h e = h \in HN$ satisfying $$ \phi (hn) = h (H \cap N ) $$ so $\phi$ is surjective.

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

$( \subset )$ If $n \in N$, then $\phi (n) = \phi ( en) = e (H \cap N ) = H \cap N$, so $$ n \in \ker ( \phi ) $$ $( \supset )$ If $hn \in \ker ( \phi)$, then from $\phi (hn) = H \cap N$ we have $hn \in ( H \cap N )$, so $$ hn \in N $$

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Part 5.

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

Since $\phi : HN \to H / ( H \cap N)$ is a homomorphism and surjective, $\phi ( NH ) = H / ( H \cap N)$. Meanwhile, since $N = \ker ( \phi )$, the following holds by the first isomorphism theorem. $$ (HN) / N \simeq H / (H \cap N) $$

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