Internal and External Group Direct Products are Isomorphic

Theorem
Let $G$ be a group whose identity is $e$.

Then $G$ is the (external) group direct product of $G_1, G_2, \ldots, G_n$ iff $G$ is the internal group direct product of $N_1, N_2, \ldots, N_n$ such that:


 * $\forall i \in \N^*_n: N_i \cong G_i$

where $\cong$ denotes (group) isomorphism.

Proof

 * Let $G$ be the external direct product of group $G_1, G_2, \ldots, G_n$.

For all $i \in \N^*_n$, let $N_i$ be defined as the set:


 * $N_i = \left\{{e}\right\} \times \cdots \times \left\{{e}\right\} \times G_i \times \left\{{e}\right\} \times \cdots \times \left\{{e}\right\}$

of elements which have entry $e$ everywhere except possibly in the $i$th co-ordinate.

It is easily checked that:


 * $(1): \quad N_i$ is isomorphic to $G_i$;
 * $(2): \quad N_i$ is a normal subgroup of $G$;
 * $(3): \quad$ Every element of $G$ has a unique expression:


 * $\left({g_1, \ldots, g_n}\right) = \left({g_1, e, \ldots, e}\right) \left({e, g_2, e \ldots e}\right) \ldots \left({e, \ldots, g_n}\right)$

as a product of elements of $N_1, \ldots, N_n$.


 * Now let $G$ be the internal group direct product of $N_1, N_2, \ldots, N_n$.

We define a mapping $\theta: G \to N_1 \times N_2 \times \cdots \times N_n$ by:


 * $\theta \left({g_1 g_2 \ldots g_n}\right) = \left({g_1, g_2, \ldots, g_n}\right)$

Using Internal Group Direct Product Commutativity, we can prove by induction that:

So $\theta$ is a homomorphism.

The fact that $\theta$ is a bijection follows from the definitions.