Isomorphism of External Direct Products

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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:

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

be external direct products of algebraic structures.


Let $\phi_k: \struct {S_k, \circ_k} \to \struct {T_k, \ast_k}$ be an isomorphism for each $k \in \closedint 1 n$.

Then:

$\phi: \struct {s_1, \ldots, s_n} \to \struct {\map {\phi_1} {s_1}, \ldots, \map {\phi_n} {s_n} }$

is an isomorphism from $\struct {S, \circ}$ to $\struct {T, \ast}$.


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.

$\blacksquare$


Sources

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