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.
== Proof ==
=== 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.
Category:Internal Group Direct Products