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

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

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

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

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 $\displaystyle \prod_{j=1}^n h_j = \prod_{j=1}^n g_j$.

Thus, each element of $G$ that can be expressed as a product of the form $\displaystyle \prod_{j=1}^n h_j$ 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 $\forall i, j \in \N^*_n: i \ne j \implies H_i \cap H_j = \left\{{e}\right\}$.


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

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

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

Then we need to show:


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

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

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

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

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

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

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

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

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

Hence the result.