# Isomorphism of External Direct Products/General Result

## Theorem

Let:

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

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

By definition of isomorphism, each $\phi_k$ is a homomorphism which is a bijection.

From Cartesian Product of Bijections is Bijection: General Result, $\phi$ is a bijection.

From Homomorphism of External Direct Products: General Result, $\phi$ is a homomorphism.

Hence the result.

$\blacksquare$