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
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 $$x y = y x$$.