Internal Direct Product Theorem/Proof 1
Theorem
The following definitions of the concept of Internal Group Direct Product are equivalent:
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 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 $G$.
Proof
From Conditions for Internal Group Direct Product it is sufficient to show that if:
- $\quad G = H \circ K$
and:
- $\quad H \cap K = \set e$
then:
- $\struct {H, \circ {\restriction_H} }$ and $\struct {H_2, \circ {\restriction_K} }$ are both normal subgroups of $\struct {G, \circ}$
- $\forall \tuple {h, k} \in H \times K: h \circ k = k \circ h$
Sufficient Condition
Let $\struct {H, \circ {\restriction_H} }$ and $\struct {H_2, \circ {\restriction_K} }$ both be normal subgroups of $\struct {G, \circ}$.
Let $x \in H$ and $y \in K$.
Then:
\(\ds x^{-1}\) | \(\in\) | \(\ds H\) | Two-Step Subgroup Test | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y \circ x^{-1} \circ y^{-1}\) | \(\in\) | \(\ds H\) | Definition of Normal Subgroup | ||||||||||
\(\ds x \circ y \circ x^{-1}\) | \(\in\) | \(\ds K\) | Definition of Normal Subgroup | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x \circ y \circ x^{-1} } \circ y^{-1}\) | \(\in\) | \(\ds K\) | Group Axiom $\text G 0$: Closure | ||||||||||
\(\, \ds \land \, \) | \(\ds x \circ \paren {y \circ x^{-1} \circ y^{-1} }\) | \(\in\) | \(\ds H\) | Group Axiom $\text G 0$: Closure | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \circ y \circ x^{-1} \circ y^{-1}\) | \(\in\) | \(\ds H \cap K\) | Group Axiom $\text G 1$: Associativity, Definition of Set Intersection | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x \circ y} \circ \paren {x^{-1} \circ y^{-1} }\) | \(=\) | \(\ds e\) | a priori: $H \cap K = \set e$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x \circ y} \circ \paren {y \circ x}^{-1}\) | \(=\) | \(\ds e\) | Inverse of Group Product | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x \circ y} \circ \paren {y \circ x}^{-1} \circ \paren {y \circ x}\) | \(=\) | \(\ds y \circ x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \circ y\) | \(=\) | \(\ds y \circ x\) |
$x$ and $y$ are arbitrary, so:
- $\forall \tuple {h, k} \in H \times K: h \circ k = k \circ h$
$\Box$
Necessary Condition
Let:
- $\forall \tuple {h, k} \in H \times K: h \circ k = k \circ h$
Let $z = G$.
We have:
\(\ds \exists h \in H, k \in K: \, \) | \(\ds z\) | \(=\) | \(\ds h \circ k\) | a priori: $G = H \circ K$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds k \circ h\) | by hypothesis | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds k \circ H\) | \(=\) | \(\ds H \circ k\) | by hypothesis | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds h \circ K\) | \(=\) | \(\ds K \circ h\) | by hypothesis |
Then we have:
\(\ds z \circ H \circ z^{-1}\) | \(=\) | \(\ds \paren {k \circ h} \circ H \circ \paren {h^{-1} \circ k^{-1} }\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds k \circ \paren {h \circ H \circ h^{-1} } \circ k^{-1}\) | Group Axiom $\text G 1$: Associativity | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds k \circ H \circ k^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds H \circ k \circ k^{-1}\) | a priori | |||||||||||
\(\ds \) | \(=\) | \(\ds H \circ e\) | Group Axiom $\text G 3$: Existence of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds H\) |
Similarly:
\(\ds z \circ K \circ z^{-1}\) | \(=\) | \(\ds \paren {h \circ k} \circ K \circ \paren {k^{-1} \circ h^{-1} }\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds h \circ \paren {k \circ K \circ k^{-1} } \circ h^{-1}\) | Group Axiom $\text G 1$: Associativity | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds h \circ K \circ h^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds K \circ h \circ h^{-1}\) | a priori | |||||||||||
\(\ds \) | \(=\) | \(\ds K \circ e\) | Group Axiom $\text G 3$: Existence of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds K\) |
Thus, by definition, $H$ and $K$ are both $\struct {H, \circ {\restriction_H} }$ and $\struct {H_2, \circ {\restriction_K} }$ are both normal subgroups of $\struct {G, \circ}$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.5$