External Direct Product Inverses
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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 $\struct {T, \circ_2}$.
Then:
- $s^{-1}$ and $t^{-1}$ are inverse elements of $s \in \struct {S, \circ_1}$ and $t \in \struct {T, \circ_2}$ respectively
- $\tuple {s^{-1}, t^{-1} }$ is an inverse element of $\tuple {s, t} \in \struct {S \times T, \circ}$.
Proof
Sufficient Condition
Let:
- $e_S$ be the identity for $\struct {S, \circ_1}$
and:
- $e_T$ be the identity for $\struct {T, \circ_2}$.
Also let:
- $s^{-1}$ be the inverse of $s \in \struct {S, \circ_1}$
and
- $t^{-1}$ be the inverse of $t \in \struct {T, \circ_2}$.
Then:
| \(\ds \tuple {s, t} \circ \tuple {s^{-1}, t^{-1} }\) | \(=\) | \(\ds \tuple {s \circ_1 s^{-1}, t \circ_2 t^{-1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \tuple {e_S, e_T}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \tuple {s^{-1} \circ_1 s, t^{-1} \circ_2 t}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \tuple {s^{-1}, t^{-1} } \circ \tuple {s, t}\) |
Thus the inverse of $\tuple {s, t}$ is $\tuple {s^{-1}, t^{-1} }$.
$\Box$
Necessary Condition
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.
$\blacksquare$