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:

\(\displaystyle H_i \cap H_j\) \(=\) \(\displaystyle \set e\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle g\) \(=\) \(\displaystyle h_i \circ h_j \circ h_i^{-1} \circ h_j^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle e\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle h_i \circ h_j \circ h_i^{-1}\) \(=\) \(\displaystyle h_j\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle h_i \circ h_j\) \(=\) \(\displaystyle h_j \circ h_i\)

$\blacksquare$


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