Equivalence of Definitions of Internal Group Direct Product
Theorem
The following definitions of the concept of Internal Group Direct Product are equivalent:
Definition $1$: Definition by Isomorphism
The group $\struct {G, \circ}$ is the internal group direct product of $H$ and $K$ if and only if 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 the (external) group direct product $\struct {H, \circ {\restriction_H} } \times \struct {K, \circ {\restriction_K} }$ onto $\struct {G, \circ}$.
Definition $2$: Definition by Subset Product
The group $\struct {G, \circ}$ is the internal group direct product of $H$ and $K$ if and only if:
- $(1): \quad \struct {H, \circ {\restriction_H} }$ and $\struct {K, \circ {\restriction_K} }$ are both normal subgroups of $\struct {G, \circ}$
- $(2): \quad G$ is the subset product of $H$ and $K$, that is: $G = H \circ K$
- $(3): \quad$ $H \cap K = \set e$ where $e$ is the identity element of $G$.
Definition $3$: Definition by Unique Expression
The group $\struct {G, \circ}$ is the internal group direct product of $H$ and $K$ if and only if:
- $(1): \quad \struct {H, \circ {\restriction_H} }$ 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 $k \in K$.
Proof
Let $e$ denote the identity element of $\struct {G, \circ}$.
$(1)$ if and only if $(2)$
This is demonstrated in the Internal Direct Product Theorem.
$\blacksquare$
$(1)$ 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 $h \circ k = g$.
$\Box$
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 $3$.
$\Box$
$(3)$ 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 $(2)$:
\(\ds G\) | \(=\) | \(\ds \set {h \circ k: h \in H, k \in K}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds H \circ K\) | Definition of Subset Product |
$\Box$
Suppose $x \in H \cap K$.
Recall that $e$ denotes the identity element of $\struct {G, \circ}$.
We have:
\(\ds x = x \circ e\) | \(:\) | \(\ds x \in H, e \in K\) | ||||||||||||
\(\ds x = e \circ x\) | \(:\) | \(\ds e \in H, x \in K\) |
Because of uniqueness of representation:
\(\ds x \circ e\) | \(=\) | \(\ds e \circ x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds e\) |
Thus:
- $H \cap K = \set e$
Thus $\struct {G, \circ}$ is the internal group direct product of $H$ and $K$ by definition $2$.
$\blacksquare$