=== Necessary Condition === Let $G$ be the external direct product of groups $G_1, G_2, \ldots, G_n$. For all $i \in \N_n$, let $N_i$ be defined as the set: :$N_i = \set e \times \cdots \times \set e \times G_i \times \set e \times \cdots \times \set e$ of elements which have entry $e$ everywhere except possibly in the $i$th co-ordinate. It remains to be 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: :$\tuple {g_1, \ldots, g_n} = \tuple {g_1, e, \ldots, e} \tuple {e, g_2, e \ldots e} \ldots \tuple {e, \ldots, g_n}$ as a product of elements of $N_1, \ldots, N_n$. ==== $N_i$ is isomorphic to $G_i$ ==== {{finish}} ==== $N_i$ is a normal subgroup of $G$ ==== {{finish}} ==== Every element of $G$ has a unique expression ==== {{finish}} === Sufficient Condition === 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: :$\map \theta {g_1 g_2 \ldots g_n} = \tuple {g_1, g_2, \ldots, g_n}$ The fact that $\theta$ is a bijection follows from the definitions. It remains to be shown that $\theta$ is a homomorphism. The proof proceeds by induction. For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition: :$\map \theta {g_1 g_2 \ldots g_n} \map \theta {h_1 h_2 \ldots h_n} = \map \theta {g_1 g_2 \ldots g_n h_1 h_2 \ldots h_n}$ ==== Basis for the Induction ==== $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = \map \theta {g_1 g_2} \map \theta {h_1 h_2}       | r = \tuple {g_1, g_2} \tuple {h_1, h_2}       | c = Definition of $\theta$ }} {{eqn | r = \tuple {g_1 h_1, g_2 h_2}       | c =  }} {{eqn | r = \map \theta {g_1 h_1 g_2 h_2}       | c = Definition of $\theta$ }} {{eqn | r = \map \theta {g_1 g_2 h_1 h_2}       | c = Internal Group Direct Product Commutativity: $h_1 g_2 = g_2 h_1$ }} {{end-eqn}} Thus $\map P 2$ is seen to hold. This is the basis for the induction. ==== Induction Hypothesis ==== Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\map \theta {g_1 g_2 \ldots g_k} \map \theta {h_1 h_2 \ldots h_k} = \map \theta {g_1 g_2 \ldots g_k h_1 h_2 \ldots h_k}$ from which it is to be shown that: :$\map \theta {g_1 g_2 \ldots g_{k + 1} } \map \theta {h_1 h_2 \ldots h_{k + 1} } = \map \theta {g_1 g_2 \ldots g_{k + 1} h_1 h_2 \ldots h_{k + 1} }$ ==== Induction Step ==== This is the induction step: {{begin-eqn}} {{eqn | o =        | r = \map \theta {g_1 g_2 \ldots g_{k + 1} } \map \theta {h_1 h_2 \ldots h_{k + 1} }       | c =  }} {{eqn | r = \map \theta {\paren {g_1 g_2 \ldots g_k} g_{k + 1} } \map \theta {\paren {h_1 h_2 \ldots h_k} h_{k + 1} }       | c = {{GroupAxiom|1}} }} {{eqn | r = \map \theta {\paren {g_1 g_2 \ldots g_k} g_{k + 1} \paren {h_1 h_2 \ldots h_k} h_{k + 1} }       | c = Basis for the Induction }} {{eqn | r = \map \theta {g_1 g_2 \ldots g_k g_{k + 1} h_1 h_2 \ldots h_k h_{k + 1} }       | c = {{GroupAxiom|1}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>1}: \map \theta {g_1 g_2 \ldots g_n} \map \theta {h_1 h_2 \ldots h_n} = \map \theta {g_1 g_2 \ldots g_n h_1 h_2 \ldots h_n}$ So $\theta$ is a homomorphism. {{qed}}