Internal Group Direct Product is Injective

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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:

$\map \phi {h_1, h_2} = h_1 h_2$


Then $\phi$ is injective if and only if:

$H_1 \cap H_2 = \set e$


General Result

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 if and only if:

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

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


Proof

Necessary Condition

Let $\phi$ be an injection.

Let $\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$.

As $\phi$ is injective, this means that:

$\tuple {h_1, h_2} = \tuple {k_1, k_2}$

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:

\(\ds h = h e\) \(:\) \(\ds h \in H_1, e \in H_2\)
\(\ds h = e h\) \(:\) \(\ds e \in H_1, h \in H_2\)


Thus we see that:

\(\ds \map \phi {h, e}\) \(=\) \(\ds \map \phi {e, h}\)
\(\ds \leadsto \ \ \) \(\ds \tuple {h, e}\) \(=\) \(\ds \tuple {e, h}\)
\(\ds \leadsto \ \ \) \(\ds h\) \(=\) \(\ds e\)


Thus:

$H_1 \cap H_2 = \set e$

$\Box$


Sufficient Condition

Let $H_1 \cap H_2 = \set e$.

Let:

$\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$

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:

$k_1^{-1} h_1 = k_2 h_2^{-1} \in H_1 \cap H_2 = \set e$

It follows that:

$h_1 = k_1, h_2 = k_2$

and thus:

$\tuple {h_1, h_2} = \tuple {k_1, k_2}$

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

$\blacksquare$