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:


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

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 (WIP) that:


 * 1) $$N_i$$ is isomorphic to $$G_i$$;
 * 2) $$N_i$$ is a normal subgroup of $$G$$;
 * 3) 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 (WIP) that:

$$ $$

So $$\theta$$ is a homomorphism.

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