# Internal Group Direct Product Commutativity/Proof 2

## 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.

$\blacksquare$