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 $\struct {G, \circ}$
- $\forall \tuple {h, k} \in H \times K: h \circ k = k \circ h$
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:
$\quad \begin{array}{r|rrrrrr} \struct {\Z_6, +_6} & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \hline \eqclass 0 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \eqclass 1 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 \\ \eqclass 2 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 \\ \eqclass 3 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 \\ \eqclass 4 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 \\ \eqclass 5 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 \\ \end{array}$
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 $\struct {\Z_3, +_3} \times \struct {\Z_2, +_2}$.