Internal Group Direct Product is Injective/General Result

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

Let $\left \langle {H_n} \right \rangle$ be a sequence of subgroups of $G$.

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

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

That is, $\left \langle {H_n} \right \rangle$ is a sequence of independent subgroups.

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

Let $\phi_n$ be an injection.

Let $\phi_n \left({\left({h_1, h_2, \ldots, h_n}\right)}\right) = \phi_n \left({\left({g_1, g_2, \ldots, g_n}\right)}\right)$.

As $\phi_n$ is injective:
 * $\left({h_1, h_2, \ldots, h_n}\right) = \left({g_1, g_2, \ldots, g_n}\right)$

and thus
 * $\forall i \in \left\{{1, 2, \ldots, n}\right\}: h_i = g_i$

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 $i, j \in \left\{{1, 2, \ldots, n}\right\}: i \ne j$.

Suppose $h \in H_i \cap H_j$.

Thus we see that:

Thus $H_i \cap H_j = \left\{{e}\right\}$.

This holds for all pairs of integers $i, j \in \left\{{1, 2, \ldots, n}\right\}$ where $i \ne j$.

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

Sufficient Condition
Let:
 * $\forall i, j \in \left\{{1, 2, \ldots, n}\right\}: i \ne j \implies H_i \cap H_j = \left\{{e}\right\}$

Let:
 * $\phi_n \left({\left({h_1, h_2, \ldots, h_n}\right)}\right) = \phi_n \left({\left({g_1, g_2, \ldots, g_n}\right)}\right)$

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 $n \in \left\{{1, 2, \ldots, n}\right\}$, let $P \left({n}\right)$ be the proposition:
 * If $\left({\forall i, j \in \left\{{1, 2, \ldots, n}\right\}: i \ne j \implies H_i \cap H_j = \left\{{e}\right\}}\right)$ then $\phi_n$ is injective.

Basis for the Induction
$P(1)$ is trivially true.

$P(2)$ is the case:
 * If $H_1 \cap H_2 = \left\{{e}\right\}$, 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 $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:


 * If $\left({\forall i, j \in \left\{{1, 2, \ldots, k}\right\}: i \ne j \implies H_i \cap H_j = \left\{{e}\right\}}\right)$ then $\phi_k$ is injective.

Then we need to show:


 * If $\left({\forall i, j \in \left\{{1, 2, \ldots, k+1}\right\}: i \ne j \implies H_i \cap H_j = \left\{{e}\right\}}\right)$ then $\phi_{k+1}$ is injective.

Induction Step
This is our induction step:

Suppose:
 * $\phi_{k+1} \left({\left({h_1, h_2, \ldots, h_{k+1}}\right)}\right) = \phi_{k+1} \left({\left({g_1, g_2, \ldots, g_{k+1}}\right)}\right)$

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

or:
 * $\displaystyle \left({\prod_{j \mathop = 1}^k h_j}\right) h_{k+1} = \left({\prod_{j \mathop = 1}^k g_j}\right) g_{k+1}$

Thus:
 * $\displaystyle g_{k+1}^{-1} h_{k+1} = \left({\prod_{j \mathop = 1}^k g_j}\right) \left({\prod_{j \mathop = 1}^k h_j}\right)^{-1}$

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

and hence:
 * $\displaystyle \left({\prod_{j \mathop = 1}^k g_j}\right) \left({\prod_{j \mathop = 1}^k h_j}\right)^{-1} \in \bigcup_{i \mathop = 1}^k H_i$

But:
 * $\displaystyle \left({\prod_{j \mathop = 1}^k g_j}\right) \left({\prod_{j \mathop = 1}^k h_j}\right)^{-1} \in \bigcup_{i \mathop = 1}^k H_i$

Thus:
 * $\displaystyle g_{k+1}^{-1} h_{k+1} = \left({\prod_{j \mathop = 1}^k g_j}\right) \left({\prod_{j \mathop = 1}^k h_j}\right)^{-1} \in H_{k+1} \cap \bigcup_{i \mathop = 1}^k H_i$

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 $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Hence the result.