Equivalence of Definitions of Internal Group Direct Product

Proof
Let $e$ denote the identity element of $\struct {G, \circ}$.

$(1)$ implies $(2)$
Let $\struct {G, \circ}$ be the internal group direct product of $H_1$ and $H_2$ by definition $1$.

Then by definition the mapping $\phi: H_1 \times H_2 \to G$ defined as:


 * $\forall h_1 \in H_1, h_2 \in H_2: \map \phi {h_1, h_2} = h_1 \circ h_2$

is a group isomorphism from $\struct {H_1, \circ {\restriction_{H_1} } } \times \struct {H_2, \circ {\restriction_{H_2} } }$ onto $\struct {G, \circ}$.

As $\phi$ is an isomorphism, it is a bijection.

As $\phi$ is a bijection, it is surjective.

Thus by definition of surjection, every element $g$ of $G$ can be expressed in the form:
 * $g = h_1 \circ h_2$

where $h_1 \in H_1$ and $h_2 \in H_2$

As $\phi$ is a bijection, it is injective.

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

Then by definition of injective:
 * $\tuple {h_1, k_1} = \tuple {h_2, k_2}$

and thus:
 * $h_1 = h_2, k_1 = k_2$

From the definition of $\phi$, this means:
 * $h_1 \circ k_1 = h_2 \circ k_2$

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

We now need to show that $H_1$ and $H_2$ are normal subgroups of $G$.

This is demonstrated in Internal Group Direct Product Isomorphism.

Thus we have shown that $\struct {G, \circ}$ is the internal group direct product of $H_1$ and $H_2$ by definition $2$.

$(2)$ implies $(3)$
Let $\struct {G, \circ}$ be the internal group direct product of $H_1$ and $H_2$ by definition $2$.

Then by definition:
 * $(1): \quad \struct {H_1, \circ {\restriction_{H_1} } }$ and $\struct {H_2, \circ {\restriction_{H_2} } }$ are both normal subgroups of $\struct {G, \circ}$


 * $(2): \quad$ every element of $G$ can be expressed uniquely in the form:
 * $g = h_1 \circ h_2$
 * where $h_1 \in H_1$ and $h_2 \in H_2$.

Criterion $(1)$ is common to both definitions.

It remains to be shown that:


 * $G$ is the subset product of $H_1$ and $H_2$, that is: $G = H_1 \circ H_2$


 * $H_1 \cap H_2 = \set e$, where $e$ is the identity element of $G$.

Indeed, from $(2)$:

Suppose $x \in H_1 \cap H_2$.

Recall that $e$ denotes the identity element of $\struct {G, \circ}$.

We have:

Because of uniqueness of representation:

Thus:
 * $H_1 \cap H_2 = \set e$

Thus $\struct {G, \circ}$ is the internal group direct product of $H_1$ and $H_2$ by definition $3$.

$(3)$ implies $(1)$
Let $\struct {G, \circ}$ be the internal group direct product of $H_1$ and $H_2$ by definition $3$.

Then by definition:
 * $(1): \quad \struct {H_1, \circ {\restriction_{H_1} } }$ and $\struct {H_2, \circ {\restriction_{H_2} } }$ are both normal subgroups of $\struct {G, \circ}$


 * $(2): \quad G$ is the subset product of $H_1$ and $H_2$, that is: $G = H_1 \circ H_2$


 * $(3): \quad$ $H_1 \cap H_2 = \set e$ where $e$ is the identity element of $G$.

Let $\phi: H_1 \times H_2 \to G$ be the mapping defined as:


 * $\forall h_1 \in H_1, h_2 \in H_2: \map \phi {h_1, h_2} = h_1 \circ h_2$

It remains to be shown that this is an isomorphism.

This is demonstrated in the Internal Direct Product Theorem.

Thus $\struct {G, \circ}$ is the internal group direct product of $H_1$ and $H_2$ by definition $1$.