Equivalence of Definitions of Internal Group Direct Product

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

$(1)$ $(2)$
This is demonstrated in the Internal Direct Product Theorem.

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

Then by definition the mapping $\phi: H \times K \to G$ defined as:


 * $\forall h \in H, k \in K: \map \phi {h, k} = h \circ k$

is a group isomorphism from $\struct {H, \circ {\restriction_H} } \times \struct {K, \circ {\restriction_K} }$ onto $\struct {G, \circ}$.

From Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection:
 * for all $g \in G$: there exists a unique $\tuple {h, k} \in H \times K$ such that $h \circ k = g$.

We now need to show that $H$ and $K$ 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 $H$ and $K$ by definition $3$.

$(3)$ implies $(2)$
Criterion $(1)$ is common to both definitions.

Let $\struct {G, \circ}$ be the internal group direct product of $H$ and $K$ by definition $3$.

Then by definition:
 * $(1): \quad \struct {H, \circ {\restriction_H} }$ and $\struct {K, \circ {\restriction_K} }$ are both normal subgroups of $\struct {G, \circ}$


 * $(2): \quad$ every element of $G$ can be expressed uniquely in the form:
 * $g = h \circ k$
 * where $h \in H$ and $k \in K$.

It remains to be shown that:


 * $G$ is the subset product of $H$ and $K$, that is: $G = H \circ K$


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

Indeed, from $(2)$:

Suppose $x \in H \cap K$.

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

We have:

Because of uniqueness of representation:

Thus:
 * $H \cap K = \set e$

Thus $\struct {G, \circ}$ is the internal group direct product of $H$ and $K$ by definition $2$.