Internal Direct Product Theorem

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 $G$.

General Result

Let $G$ be a group whose identity is $e$.

Let $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of subgroups of $G$.

Then $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ if and only if:

$(1): \quad G = H_1 H_2 \cdots H_n$

$(2): \quad \sequence {H_k}_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups

$(3): \quad \forall k \in \set {1, 2, \ldots, n}: H_k \lhd G$

where $H_k \lhd G$ denotes that $H_k$ is a normal subgroup of $G$.

Proof 1

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 $\struct {G, \circ}$.

$\blacksquare$

Proof 2

Sufficient Condition

Let $\phi: H \times K \to G$ be the mapping defined as:

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

Let $\phi$ be an isomorphism.

$(1): \quad$ From Codomain of Internal Direct Isomorphism is Subset Product of Factors, $G = H \circ K$.

$(2): \quad$ From Internal Group Direct Product is Injective, $H$ and $K$ are independent subgroups of $G$.

$(3): \quad$ From Internal Group Direct Product Isomorphism, $H$ and $K$ are normal subgroups of $G$.

$\Box$

Necessary Condition

Let $\phi: H \times K \to G$ be the mapping defined as:

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

Suppose the three conditions hold.

$(1): \quad$ From $G = H \circ K$, $\phi$ is surjective.

$(2): \quad$ From Internal Group Direct Product is Injective, $\phi$ is injective.

$(3): \quad$ From Internal Group Direct Product of Normal Subgroups, $\phi$ is a group homomorphism.

Putting these together, we see that $\phi$ is a bijective homomorphism, and therefore an isomorphism.

So by definition, $G$ is the internal group direct product of $H$ and $K$.

$\blacksquare$

Proof 3

A specific instance of the general result, with $n = 2$.

$\blacksquare$

Examples

Symmetry Group of Rectangle

Consider the symmetry group of the rectangle $D_2$:

Let $\RR = ABCD$ be a (non-square) rectangle.

The various symmetry mappings of $\RR$ are:

The identity mapping $e$

The rotation $r$ (in either direction) of $180^\circ$

The reflections $h$ and $v$ in the indicated axes.

The symmetries of $\RR$ form the dihedral group $D_2$.

Let $H := \set {e, r}$.

Let $K := \set {e, h}$.

Then $H$ and $K$ are subgroups of $D_2$ which fulfil the conditions of the Internal Direct Product Theorem, as:

$r \circ h = v = h \circ r$

Thus $D_2$ is the internal group direct product of $H$ and $K$.

Both $H$ and $K$ are isomorphic to $\struct {\Z_2, +_2}$, the additive group of integers modulo $2$.

Hence by Isomorphism of External Direct Products:

$D_2$ is isomorphic to $\struct {\Z_2, +_2} \times \struct {\Z_2, +_2}$.

Additive Group of Integers Modulo $6$

Consider the additive group of integers modulo $6$ $\struct {\Z_6, \times_6}$, illustrated by Cayley Table:

$\begin{array}{r|rrrrrr} \struct {\Z_6, +_6} & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \hline \eqclass 0 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \eqclass 1 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 \\ \eqclass 2 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 \\ \eqclass 3 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 \\ \eqclass 4 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 \\ \eqclass 5 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 \\ \end{array}$

Let $H := \set {0, 2, 4}$.

Let $K := \set {0, 3}$.

We have that:

$H +_6 K = \struct {\Z_6, +_6}$

and:

$H \cap K = \set 0$

Hence $H$ and $K$ are subgroups of $\struct {\Z_6, +_6}$ which fulfil the conditions of the Internal Direct Product Theorem.

Thus $\struct {\Z_6, +_6}$ is the internal group direct product of $H$ and $K$.

Because:

$H$ is isomorphic to $\struct {\Z_3, +_3}$

$K$ is isomorphic to $\struct {\Z_2, +_2}$

it follows by Isomorphism of External Direct Products that:

$\struct {\Z_6, +_6}$ is isomorphic to $\struct {\Z_3, +_3} \times \struct {\Z_2, +_2}$.