Isomorphism of External Direct Products

Theorem

Let:

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

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

$\left({\phi_1 \times \phi_2}\right) \left({\left({x, y}\right)}\right) = \left({\phi_1 \left({x}\right), \phi_2 \left({y}\right)}\right)$

is an isomorphism from $\left({S_1 \times S_2, \circ}\right)$ to $\left({T_1 \times T_2, *}\right)$.

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 $\phi_1 \times \phi_2: \left({S_1 \times S_2, \circ}\right) \to \left({T_1 \times T_2, *}\right)$ 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.

$\blacksquare$