Internal Group Direct Product Commutativity/Proof 1

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Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $H$ and $K$ be subgroups of $G$.

Let $\struct {G, \circ}$ be the internal group direct product of $H$ and $K$.


Then:

$\forall h \in H, k \in K: h \circ k = k \circ h$


Proof

Let $G$ be the internal group direct product of $H$ and $K$.

Then by definition the mapping:

$\phi: H \times K \to G: \map \phi {h, k} = h \circ k$

is a (group) isomorphism from the (external) direct product $\struct {H, \circ \restriction_H} \times \struct {K, \circ \restriction_K}$ onto $\struct {G, \circ}$.

Let the symbol $\circ$ also be used for the operation induced on $H \times K$ by $\circ \restriction_H$ and $\circ \restriction_K$.


Let $h \in H, k \in K$.

Then:

\(\ds \tuple {e, k} \circ \tuple {h, e}\) \(=\) \(\ds \tuple {e \circ h, k \circ e}\) Definition of External Direct Product
\(\ds \) \(=\) \(\ds \tuple {h, k}\) Definition of Identity Element
\(\ds \leadsto \ \ \) \(\ds h \circ k\) \(=\) \(\ds \map \phi {h, k}\) Definition of Internal Direct Product
\(\ds \) \(=\) \(\ds \map \phi {\tuple {e, k} \circ \tuple {h, e} }\) a priori
\(\ds \) \(=\) \(\ds \map \phi {e, k} \circ \map \phi {h, e}\) as $\phi$ is a homomorphism
\(\ds \) \(=\) \(\ds \paren {e \circ k} \circ \paren {h \circ e}\) Definition of Internal Direct Product
\(\ds \) \(=\) \(\ds k \circ h\) Definition of Identity Element

$\blacksquare$


Sources

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