# Internal Group Direct Product Commutativity

## 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:

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

### General Result

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

Let $h_i$ and $h_j$ be elements of $H_i$ and $H_j$ respectively, $i \ne j$.

Then $h_i \circ h_j = h_j \circ h_i$.

## Proof 1

Let $G$ be the internal group direct product of $H_1$ and $H_2$.

Then by definition the mapping:

$C: H_1 \times H_2 \to G: \map C {h_1, h_2} = h_1 \circ h_2$

is a (group) isomorphism from the cartesian product $\struct {H_1, \circ \restriction_{H_1} } \times \struct {H_2, \circ \restriction_{H_2} }$ onto $\struct {G, \circ}$.

Let the symbol $\circ$ also be used for the operation induced on $H_1 \times H_2$ by $\circ \restriction_{H_1}$ and $\circ \restriction_{H_2}$.

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

Then:

 $\displaystyle \tuple {e, h_2} \circ \tuple {h_1, e}$ $=$ $\displaystyle \tuple {e \circ h_1, h_2 \circ e}$ Definition of Internal Direct Product $\displaystyle$ $=$ $\displaystyle \tuple {h_1, h_2}$ Definition of Identity Element $\displaystyle \leadsto \ \$ $\displaystyle h_1 \circ h_2$ $=$ $\displaystyle \map C {h_1, h_2}$ $\displaystyle$ $=$ $\displaystyle \map C {\tuple {e, h_2} \circ \tuple {h_1, e} }$ $\displaystyle$ $=$ $\displaystyle \map C {e, h_2} \circ \map C {h_1, e}$ $\displaystyle$ $=$ $\displaystyle \paren {e \circ h_2} \circ \paren {h_1 \circ e}$ $\displaystyle$ $=$ $\displaystyle h_2 \circ h_1$

$\blacksquare$

## Proof 2

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

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

We have that:

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

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 $G \, 1$: 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 $G \, 1$: 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.

$\blacksquare$