=== Necessary Condition === Let $\displaystyle \phi_n: \prod_{j \mathop = 1}^n H_j \to G$ be a mapping defined by: :$\displaystyle \map {\phi_n} {h_1, h_2, \ldots, h_n} = \prod_{j \mathop = 1}^n h_j$ Let $\phi_n$ be an injection. Let $\map {\phi_n} {h_1, h_2, \ldots, h_n} = \map {\phi_n} {g_1, g_2, \ldots, g_n}$. As $\phi_n$ is injective: :$\tuple {h_1, h_2, \ldots, h_n} = \tuple {g_1, g_2, \ldots, g_n}$ and thus :$\forall i \in \set {1, 2, \ldots, n}: h_i = g_i$ From the definition of $\phi_n$: : $\displaystyle \prod_{j \mathop = 1}^n h_j = \prod_{j \mathop = 1}^n g_j$ Thus, each element of $G$ that can be expressed as a product of the form $\displaystyle \prod_{j \mathop = 1}^n h_j$ can be thus expressed ''uniquely''. Let $i, j \in \set {1, 2, \ldots, n}: i \ne j$. Suppose $h \in H_i \cap H_j$. {{begin-eqn}} {{eqn | l = h = h e       | o = :       | r = h \in H_i, e \in H_j       | c =  }} {{eqn | l = h = e h       | o = :       | r = e \in H_i, h \in H_j       | c =  }} {{end-eqn}} Thus we see that: {{begin-eqn}} {{eqn | l = \map {\phi_n} {e, e, \ldots, h, \ldots, e, \ldots, e, e}       | r = \map {\phi_n} {e, e, \ldots, e, \ldots, h, \ldots, e, e}       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {e, e, \ldots, h, \ldots, e, \ldots, e, e}       | r = \tuple {e, e, \ldots, e, \ldots, h, \ldots, e, e}       | c =  }} {{eqn | ll= \leadsto       | l = h       | r = e       | c =  }} {{end-eqn}} Thus $H_i \cap H_j = \set e$. This holds for ''all'' pairs of integers $i, j \in \set {1, 2, \ldots, n}$ where $i \ne j$. Thus: :$\forall i, j \in \set {1, 2, \ldots, n}: i \ne j \implies H_i \cap H_j = \set e$ {{qed|lemma}} === Sufficient Condition === Let: :$\forall i, j \in \set {1, 2, \ldots, n}: i \ne j \implies H_i \cap H_j = \set e$ Let: :$\map {\phi_n} {h_1, h_2, \ldots, h_n} = \map {\phi_n} {g_1, g_2, \ldots, g_n}$ Then: :$\displaystyle \prod_{j \mathop = 1}^n h_j = \prod_{j \mathop = 1}^n g_j: h_j, g_j \in H_j$ For all $n \in \set {1, 2, \ldots, n}$, let $\map P n$ be the proposition: :If $\paren {\forall i, j \in \set {1, 2, \ldots, n}: i \ne j \implies H_i \cap H_j = \set e}$ then $\phi_n$ is injective. ==== Basis for the Induction ==== $\map P 1$ is trivially true. $\map P 2$ is the case: :If $H_1 \cap H_2 = \set e$, then $\phi$ is injective which has been proved above. This is our basis for the induction. ==== Induction Hypothesis ==== Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :If $\paren {\forall i, j \in \set {1, 2, \ldots, k}: i \ne j \implies H_i \cap H_j = \set e}$ then $\phi_k$ is injective. Then we need to show: :If $\paren {\forall i, j \in \set {1, 2, \ldots, k + 1}: i \ne j \implies H_i \cap H_j = \set e}$ then $\phi_{k + 1}$ is injective. ==== Induction Step ==== This is our induction step: Suppose: :$\map {\phi_{k + 1} } {h_1, h_2, \ldots, h_{k + 1} } = \map {\phi_{k + 1} } {g_1, g_2, \ldots, g_{k + 1} }$ Then: :$\displaystyle \prod_{j \mathop = 1}^{k + 1} h_j = \prod_{j \mathop = 1}^{k + 1} g_j$ or: :$\displaystyle \paren {\prod_{j \mathop = 1}^k h_j} h_{k + 1} = \paren {\prod_{j \mathop = 1}^k g_j} g_{k + 1}$ Thus: :$\displaystyle g_{k + 1}^{-1} h_{k + 1} = \paren {\prod_{j \mathop = 1}^k g_j} \paren {\prod_{j \mathop = 1}^k h_j}^{-1}$ Then: :$g_{k + 1}^{-1} h_{k + 1} \in H_{k + 1}$ and hence: :$\displaystyle \paren {\prod_{j \mathop = 1}^k g_j} \paren {\prod_{j \mathop = 1}^k h_j}^{-1} \in \bigcup_{i \mathop = 1}^k H_i$ But: :$\displaystyle \paren {\prod_{j \mathop = 1}^k g_j} \paren {\prod_{j \mathop = 1}^k h_j}^{-1} \in \bigcup_{i \mathop = 1}^k H_i$ Thus: :$\displaystyle g_{k + 1}^{-1} h_{k + 1} = \paren {\prod_{j \mathop = 1}^k g_j} \paren {\prod_{j \mathop = 1}^k h_j}^{-1} \in H_{k + 1} \cap \bigcup_{i \mathop = 1}^k H_i$ But from the induction hypothesis: :$\displaystyle \bigcup_{i \mathop = 1}^k H_i = e$ So: :$\displaystyle \bigcup_{i \mathop = 1}^k H_i \cap H_{k + 1} = \bigcup_{i \mathop = 1}^{k + 1} H_i = e$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Hence the result. {{qed}} Category:Internal Group Direct Products iiho8j7ir3ky33t4z257r1bpfhwlnvz