External Direct Product Inverses

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let:

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

and:

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


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


General Result

Let $\displaystyle \left({S, \circ}\right) = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\left({S_1, \circ_1}\right), \left({S_2, \circ_2}\right), \ldots, \left({S_n, \circ_n}\right)$.

Let $\left({x_1, x_2, \ldots, x_n}\right) \in S$.


Let $y_k$ be an inverse of $x_k$ in $\left({S_k, \circ_k}\right)$ for each of $k \in \N^*_n$.

Then $\left({y_1, y_2, \ldots, y_n}\right)$ is the inverse of $\left({x_1, x_2, \ldots, x_n}\right) \in S$ in $\left({S, \circ}\right)$.


Proof

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:

\(\displaystyle \tuple {s, t} \circ \tuple {s^{-1}, t^{-1} }\) \(=\) \(\displaystyle \tuple {s \circ_1 s^{-1}, t \circ_2 t^{-1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {e_S, e_T}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {s^{-1} \circ_1 s, t^{-1} \circ_2 t}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {s^{-1}, t^{-1} } \circ \tuple {s, t}\)


Thus the inverse of $\tuple {s, t}$ is $\tuple {s^{-1}, t^{-1} }$.

$\blacksquare$


Sources