Inverses in Group Direct Product/Proof 1

From ProofWiki
Jump to navigation Jump to search

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 $\tuple {g^{-1}, h^{-1} }$.

$\blacksquare$