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 $\struct {T_1 \times T_2, *}$.

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.