Internal Group Direct Product Commutativity/General Result
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Theorem
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
Let $g = h_i \circ h_j \circ h_i^{-1} \circ h_j^{-1}$.
From the Internal Direct Product Theorem: General Result, $H_i$ and $H_j$ are normal in $G$.
Hence $h_i \circ h_j \circ h_i^{-1} \in H_j$ and thus $g \in H_j$.
Similarly, $g \in H_i$ and thus $g \in H_i \cap H_j$.
But:
\(\ds H_i \cap H_j\) | \(=\) | \(\ds \set e\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds g\) | \(=\) | \(\ds h_i \circ h_j \circ h_i^{-1} \circ h_j^{-1}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds e\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds h_i \circ h_j \circ h_i^{-1}\) | \(=\) | \(\ds h_j\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds h_i \circ h_j\) | \(=\) | \(\ds h_j \circ h_i\) |
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $13$: Direct products: Corollary $13.6$