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 $x y = y x$.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 13$: Theorem $13.4$
• John F. Humphreys: A Course in Group Theory (1996): $\S 13$: Corollary $13.6$