Internal Group Direct Product Commutativity

Theorem
Let $G$ be a group.

Let $G$ be the internal group direct product of $G_1, G_2, \ldots, G_n$.

Let $x$ and $y$ be elements of $G_i$ and $G_j$ respectively, $i \ne j$.

Then $x y = y x$.

Proof 1
Let $g = x y x^{-1} y^{-1}$.

From the Internal Direct Product Theorem, $G_i$ and $G_j$ are normal in $G$.

Hence $x y x^{-1} \in G_j$ and thus $g \in G_j$.

Similarly, $g \in G_i$ and thus $g \in G_i \cap G_j$.

But $G_i \cap G_j = \left\{{e}\right\}$ so $g = x y x^{-1} y^{-1} = e$ and thus $x y = y x$.

Proof 2
Let $G$ be the internal group direct product of $H_1$ and $H_2$.

Then by definition the mapping:
 * $C: S_1 \times S_2 \to S: C \left({\left({s_1, s_2}\right)}\right) = s_1 \circ s_2$

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

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

$(1): \quad \forall h_1 \in H_1, h_2 \in H_2: h_1 \circ h_2 = h_2 \circ h_1$:

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

Then:

Thus $(1)$ holds.