# External Direct Product Inverses

## 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.

$\blacksquare$