External Direct Product Inverses/Necessary Condition

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Theorem

Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.

Let $\tuple {s^{-1}, t^{-1} }$ be an inverse of $\tuple {s, t} \in \struct {S \times T, \circ}$.

Then:

$s^{-1}$ is an inverse of $s \in \struct {S, \circ_1}$

and:

$t^{-1}$ is an inverse of $t \in \struct {T, \circ_2}$.


Proof

Let $\tuple {e_S, e_T}$ be the identity element of $\struct {S \times T, \circ}$.

Let $\tuple {s^{-1}, t^{-1} }$ be an inverse element of $\tuple {s, t} \in \struct {S \times T, \circ}$.

Then we have:

\(\ds \tuple {s, t} \circ \tuple {s^{-1}, t^{-1} }\) \(=\) \(\ds \tuple {e_S, e_T}\) Definition of Inverse Element
\(\ds \leadsto \ \ \) \(\ds \tuple {s \circ_1 s^{-1}, t \circ_2 t^{-1} }\) \(=\) \(\ds \tuple {e_S, e_T}\) Definition of External Direct Product
\(\ds \leadsto \ \ \) \(\ds \forall s \in S, t \in T: \, \) \(\ds s \circ_1 s^{-1}\) \(=\) \(\ds e_S\) Equality of Ordered Pairs
\(\, \ds \land \, \) \(\ds t \circ_2 t^{-1}\) \(=\) \(\ds e_T\)

and:

\(\ds \tuple {s^{-1}, t^{-1} } \circ \tuple {s, t}\) \(=\) \(\ds \tuple {e_S, e_T}\) Definition of Inverse Element
\(\ds \leadsto \ \ \) \(\ds \tuple {s^{-1} \circ_1 s, t^{-1} \circ_2 t}\) \(=\) \(\ds \tuple {e_S, e_T}\) Definition of External Direct Product
\(\ds \leadsto \ \ \) \(\ds s^{-1} \circ_1 s\) \(=\) \(\ds e_S\) Equality of Ordered Pairs
\(\, \ds \land \, \) \(\ds t^{-1} \circ_2 t\) \(=\) \(\ds e_T\)

Thus $s^{-1}$ and $t^{-1}$ are inverse elements of $s \in \struct {S, \circ_1}$ and $t \in \struct {T, \circ_2}$ respectively.

$\blacksquare$


Also see


Sources

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