# 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 \ \$ $\displaystyle h_i \circ h_j$ $=$ $\displaystyle h_j \circ h_i$

$\blacksquare$