# Isomorphism of External Direct Products

## Theorem

Let:

$\struct {S_1 \times S_2, \circ}$ be the external direct product of two algebraic structures $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$
$\struct {T_1 \times T_2, *}$ be the external direct product of two algebraic structures $\struct {T_1, *_1}$ and $\struct {T_2, *_2}$
$\phi_1$ be an isomorphism from $\struct {S_1, \circ_1}$ onto $\struct {T_1, *_1}$
$\phi_2$ be an isomorphism from $\struct {S_2, \circ_2}$ onto $\struct {T_2, *_2}$.

Then the mapping $\phi_1 \times \phi_2: \struct {S_1 \times S_2, \circ} \to \struct {T_1 \times T_2, *}$ defined as:

$\map {\paren {\phi_1 \times \phi_2} } {x, y} = \tuple {\map {\phi_1} x, \map {\phi_2} y}$

is an isomorphism from $\struct {S_1 \times S_2, \circ}$ to $\struct {T_1 \times T_2, *}$.

### General Result

Let:

$(1): \quad \displaystyle \left({S, \circ}\right) = \prod_{k \mathop = 1}^n S_k = \left({S_1, \circ_1}\right) \times \left({S_2, \circ_2}\right) \times \cdots \times \left({S_n, \circ_n}\right)$
$(2): \quad \displaystyle \left({T, \ast}\right) = \prod_{k \mathop = 1}^n T_k = \left({T_1, \ast_1}\right) \times \left({T_2, \ast_2}\right) \times \cdots \times \left({T_n, \ast_n}\right)$

Let $\phi_k: \left({S_k, \circ_k}\right) \to \left({T_k, \ast_k}\right)$ be an isomorphism for each $k \in \left[{1 \,.\,.\, n}\right]$.

Then:

$\phi: \left({s_1, \ldots, s_n}\right) \to \left({\phi_1 \left({s_1}\right), \ldots, \phi_n \left({s_n}\right)}\right)$

is an isomorphism from $\left({S, \circ}\right)$ to $\left({T, \ast}\right)$.

## Proof

From Homomorphism of External Direct Products, we have that $\phi_1 \times \phi_2: \struct {S_1 \times S_2, \circ} \to \struct {T_1 \times T_2, *}$ is a homomorphism.

From Cartesian Product of Bijections is Bijection, we have that $\phi_1 \times \phi_2$ is a bijection.

Thus $\phi_1 \times \phi_2$ is a bijective homomorphism, and so an isomorphism.

$\blacksquare$