Internal Group Direct Product is Injective

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

Let $$H_1, H_2$$ be subgroups of $$G$$.

Let $$\phi: H_1 \times H_2 \to G$$ be a mapping defined by $$\phi \left({h_1, h_2}\right) = h_1 h_2$$.

Then $$\phi$$ is injective iff $$H_1 \cap H_2 = \left\{{e}\right\}$$.

Generalized Result
Let $$G$$ be a group whose identity is $$e$$.

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

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

Then $$\phi_n$$ is injective iff $$\forall i, j \in \N^*_n: i \ne j \Longrightarrow H_i \cap H_j = \left\{{e}\right\}$$.

That is, iff every pair of distinct elements of $$\left \langle {H_n} \right \rangle$$ intersect only at $$e$$.

Proof

 * First we show that if $$\phi$$ is injective, then $$H_1 \cap H_2 = \left\{{e}\right\}$$.

So, suppose $$\phi$$ is an injection.

Let $$\phi \left({\left({h_1, h_2}\right)}\right) = \phi \left({\left({k_1, k_2}\right)}\right)$$.

As $$\phi$$ is injective, this means that $$\left({h_1, h_2}\right) = \left({k_1, k_2}\right)$$, and thus $$h_1 = k_1, h_2 = k_2$$.

From the definition of $$\phi$$, this means $$h_1 h_2 = k_1 k_2$$.

Thus, each element of $$G$$ that can be expressed as a product of the form $$h_1 h_2$$ can be thus expressed uniquely.

Now, suppose $$h \in H_1 \cap H_2$$. We have:

$$ $$

Thus we see that:

$$ $$ $$

Thus $$H_1 \cap H_2 = \left\{{e}\right\}$$.


 * Now, let $$H_1 \cap H_2 = \left\{{e}\right\}$$.

Suppose $$\phi \left({\left({h_1, h_2}\right)}\right) = \phi \left({\left({k_1, k_2}\right)}\right)$$.

Then $$h_1 h_2 = k_1 k_2: h_1, k_1 \in H_1, h_2, k_2 \in H_2$$.

Thus $$k_1^{-1} h_1 = k_2 h_2^{-1}$$.

But $$k_1^{-1} h_1 \in H_1$$ and $$k_2 h_2^{-1} \in H_2$$.

As they are equal, we have $$k_1^{-1} h_1 = k_2 h_2^{-1} \in H_1 \cap H_2 = \left\{{e}\right\}$$.

It follows that $$h_1 = k_1, h_2 = k_2$$ and thus $$\left({h_1, h_2}\right) = \left({k_1, k_2}\right)$$.

Thus $$\phi$$ is injective and the result follows.

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

So, suppose $$\phi_n$$ is 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, this means that $$\left({h_1, h_2, \ldots, h_n}\right) = \left({g_1, g_2, \ldots, g_n}\right)$$, and thus $$\forall i \in N^*_n: h_i = g_i$$.

From the definition of $$\phi_n$$, this means $$\prod_{j=1}^n h_j = \prod_{j=1}^n g_j$$.

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

Let $$i, j \in N^*_n: i \ne j$$.

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 $$i, j \in \N^*_n$$ where $$i \ne j$$.

Thus $$\forall i, j \in \N^*_n: i \ne j \Longrightarrow H_i \cap H_j = \left\{{e}\right\}$$.


 * Now, suppose that $$\forall i, j \in \N^*_n: i \ne j \Longrightarrow H_i \cap H_j = \left\{{e}\right\}$$.

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

Then $$\prod_{j=1}^n h_j = \prod_{j=1}^n g_j: h_j, g_j \in H_j$$.

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition:
 * If $$\left({\forall i, j \in \N^*_n: i \ne j \Longrightarrow H_i \cap H_j = \left\{{e}\right\}}\right)$$ then $$\phi_n$$ is injective.


 * $$P(1)$$ is trivially true.

Basis for the Induction

 * $$P(2)$$ 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 $$\left({\forall i, j \in \N^*_k: i \ne j \Longrightarrow H_i \cap H_j = \left\{{e}\right\}}\right)$$ then $$\phi_k$$ is injective.

Then we need to show:


 * If $$\left({\forall i, j \in \N^*_{k+1}: i \ne j \Longrightarrow H_i \cap H_j = \left\{{e}\right\}}\right)$$ 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 $$\prod_{j=1}^{k+1} h_j = \prod_{j=1}^{k+1} g_j$$ or $$\left({\prod_{j=1}^{k} h_j}\right) h_{k+1} = \left({\prod_{j=1}^{k} g_j}\right) g_{k+1}$$.

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

Then $$g_{k+1}^{-1} h_{k+1} \in H_{k+1}$$ and $$\left({\prod_{j=1}^{k} g_j}\right) \left({\prod_{j=1}^{k} h_j}\right)^{-1} \in \bigcup_{i=1}^k H_i$$.

Thus

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

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

But from the induction hypothesis, $$\bigcup_{i=1}^k H_i = e$$.

So $$\bigcup_{i=1}^k H_i \cap H_{k+1} = \bigcup_{i=1}^{k+1} H_i = e$$.

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

Hence the result.