Inverses in Group Direct Product/Proof 1
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Theorem
Let $\struct {G \times H, \circ}$ be the group direct product of the two groups $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Let $g^{-1}$ be an inverse of $g \in \struct {G, \circ_1}$.
Let $h^{-1}$ be an inverse of $h \in \struct {H, \circ_2}$.
Then $\tuple {g^{-1}, h^{-1} }$ is the inverse of $\tuple {g, h} \in \struct {G \times H, \circ}$.
Proof
Let $e_G$ be the identity for $\struct {G, \circ_1}$
Let $e_H$ be the identity for $\struct {H, \circ_2}$.
Then:
\(\ds \tuple {g, h} \circ \tuple {g^{-1}, h^{-1} }\) | \(=\) | \(\ds \tuple {g \circ_1 g^{-1}, h \circ_2 h^{-1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {e_G, e_H}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {g^{-1} \circ_1 g, h^{-1} \circ_2 h}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {g^{-1}, h^{-1} } \circ \tuple {g, h}\) |
So the inverse of $\tuple {g, h}$ is $\tuple {g^{-1}, h^{-1} }$.
$\blacksquare$