Internal Group Direct Product is Injective/General Result

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

Let $\sequence {H_n}$ be a sequence of subgroups of $G$.

Let $\ds \phi_n: \prod_{j \mathop = 1}^n H_j \to G$ be a mapping defined by:
 * $\ds \map {\phi_n} {h_1, h_2, \ldots, h_n} = \prod_{j \mathop = 1}^n h_j$

Then $\phi_n$ is injective :
 * $\forall i, j \in \set {1, 2, \ldots, n}: i \ne j \implies H_i \cap H_j = \set e$

That is, $\sequence {H_n}$ is a sequence of independent subgroups.

Necessary Condition
Let $\ds \phi_n: \prod_{j \mathop = 1}^n H_j \to G$ be a mapping defined by:
 * $\ds \map {\phi_n} {h_1, h_2, \ldots, h_n} = \prod_{j \mathop = 1}^n h_j$

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 $\phi_n$:
 * $\ds \prod_{j \mathop = 1}^n h_j = \prod_{j \mathop = 1}^n g_j$

Thus, each element of $G$ that can be expressed as a product of the form $\ds \prod_{j \mathop = 1}^n h_j$ can be thus expressed uniquely.

Let $i, j \in \set {1, 2, \ldots, n}: i \ne j$.

Suppose $h \in H_i \cap H_j$.

Thus we see that:

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:
 * $\forall i, j \in \set {1, 2, \ldots, n}: i \ne j \implies H_i \cap H_j = \set e$

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:
 * $\ds \prod_{j \mathop = 1}^n h_j = \prod_{j \mathop = 1}^n g_j: h_j, g_j \in H_j$

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:
 * $\ds \prod_{j \mathop = 1}^{k + 1} h_j = \prod_{j \mathop = 1}^{k + 1} g_j$

or:
 * $\ds \paren {\prod_{j \mathop = 1}^k h_j} h_{k + 1} = \paren {\prod_{j \mathop = 1}^k g_j} g_{k + 1}$

Thus:
 * $\ds g_{k + 1}^{-1} h_{k + 1} = \paren {\prod_{j \mathop = 1}^k g_j} \paren {\prod_{j \mathop = 1}^k h_j}^{-1}$

Then:
 * $g_{k + 1}^{-1} h_{k + 1} \in H_{k + 1}$

and hence:
 * $\ds \paren {\prod_{j \mathop = 1}^k g_j} \paren {\prod_{j \mathop = 1}^k h_j}^{-1} \in \bigcup_{i \mathop = 1}^k H_i$

But:
 * $\ds \paren {\prod_{j \mathop = 1}^k g_j} \paren {\prod_{j \mathop = 1}^k h_j}^{-1} \in \bigcup_{i \mathop = 1}^k H_i$

Thus:
 * $\ds g_{k + 1}^{-1} h_{k + 1} = \paren {\prod_{j \mathop = 1}^k g_j} \paren {\prod_{j \mathop = 1}^k h_j}^{-1} \in H_{k + 1} \cap \bigcup_{i \mathop = 1}^k H_i$

But from the induction hypothesis:
 * $\ds \bigcup_{i \mathop = 1}^k H_i = e$

So:
 * $\ds \bigcup_{i \mathop = 1}^k H_i \cap H_{k + 1} = \bigcup_{i \mathop = 1}^{k + 1} H_i = e$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Hence the result.