Internal Group Direct Product Commutativity/Proof 2

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Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $H_1, H_2 \le G$.

Let $\struct {G, \circ}$ be the internal group direct product of $H_1$ and $H_2$.


Then:

$\forall h_1 \in H_1, h_2 \in H_2: h_1 \circ h_2 = h_2 \circ h_1$


Proof

Let $\sqbrk {x, y}$ denote the commutator of $x, y \in G$:

$\sqbrk {x, y} := x^{-1} y^{-1} x y$


We have that:

\(\text {(1)}: \quad\) \(\displaystyle y x \sqbrk {x, y}\) \(=\) \(\displaystyle y x x^{-1} y^{-1} x y\) Definition of Commutator of Group Elements
\(\displaystyle \) \(=\) \(\displaystyle y y^{-1} x y\) Group Axiom $\text G 3$: Existence of Inverse Element
\(\displaystyle \) \(=\) \(\displaystyle x y\) Group Axiom $\text G 3$: Existence of Inverse Element


Let $h_1 \in H_1$, $h_2 \in H_2$.


We have:

\(\displaystyle \sqbrk {h_1, h_2}\) \(=\) \(\displaystyle {h_1}^{-1} {h_2}^{-1} h_1 h_2\)
\(\displaystyle \) \(=\) \(\displaystyle {h_1}^{-1} \paren { {h_2}^{-1} h_1 h_2}\) Group Axiom $\text G 1$: Associativity
\(\displaystyle \) \(\in\) \(\displaystyle {h_1}^{-1} H_1\) Definition of Normal Subgroup
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqbrk {h_1, h_2}\) \(\in\) \(\displaystyle H_1\) as ${h_1}^{-1} H_1 = H_1$


and:

\(\displaystyle \sqbrk {h_1, h_2}\) \(=\) \(\displaystyle {h_1}^{-1} {h_2}^{-1} h_1 h_2\)
\(\displaystyle \) \(=\) \(\displaystyle \paren { {h_1}^{-1} {h_2}^{-1} h_1} h_2\) Group Axiom $\text G 1$: Associativity
\(\displaystyle \) \(\in\) \(\displaystyle H_2 h_2\) Definition of Normal Subgroup
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqbrk {h_1, h_2}\) \(\in\) \(\displaystyle H_2\) as $H_2 h_2 = H_2$


Thus:

$\sqbrk {h_1, h_2} \in H_1 \cap H_2$

But as $H_1 \cap H_2 = \set e$, it follows that:

$\sqbrk {h_1, h_2} = e$


It follows from Commutator is Identity iff Elements Commute that:

$h_1 h_2 = h_2 h_1$

and the result follows.

$\blacksquare$


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