Internal Direct Product Theorem/Proof 1

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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 $\struct {G, \circ}$

if and only if

$\forall \tuple {h, k} \in H \times K: h \circ k = k \circ h$


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

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