Internal Group Direct Product Commutativity
Theorem
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
Let $\struct {G, \circ}$ be the internal group direct product of $H$ and $K$.
Then:
- $\forall h \in H, k \in K: h \circ k = k \circ h$
General Result
Let $\struct {G, \circ}$ be the internal group direct product of $H_1, H_2, \ldots, H_n$.
Let $h_i$ and $h_j$ be elements of $H_i$ and $H_j$ respectively, $i \ne j$.
Then $h_i \circ h_j = h_j \circ h_i$.
Proof 1
Let $G$ be the internal group direct product of $H$ and $K$.
Then by definition the mapping:
- $\phi: H \times K \to G: \map \phi {h, k} = h \circ k$
is a (group) isomorphism from the (external) direct product $\struct {H, \circ \restriction_H} \times \struct {K, \circ \restriction_K}$ onto $\struct {G, \circ}$.
Let the symbol $\circ$ also be used for the operation induced on $H \times K$ by $\circ \restriction_H$ and $\circ \restriction_K$.
Let $h \in H, k \in K$.
Then:
| \(\ds \tuple {e, k} \circ \tuple {h, e}\) | \(=\) | \(\ds \tuple {e \circ h, k \circ e}\) | Definition of External Direct Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \tuple {h, k}\) | Definition of Identity Element | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds h \circ k\) | \(=\) | \(\ds \map \phi {h, k}\) | Definition of Internal Direct Product | ||||||||||
| \(\ds \) | \(=\) | \(\ds \map \phi {\tuple {e, k} \circ \tuple {h, e} }\) | a priori | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \phi {e, k} \circ \map \phi {h, e}\) | as $\phi$ is a homomorphism | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {e \circ k} \circ \paren {h \circ e}\) | Definition of Internal Direct Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds k \circ h\) | Definition of Identity Element |
$\blacksquare$
Proof 2
Let $\sqbrk {x, y}$ denote the commutator of $x, y \in G$:
- $\sqbrk {x, y} := x^{-1} y^{-1} x y$
We have that:
| \(\text {(1)}: \quad\) | \(\ds y x \sqbrk {x, y}\) | \(=\) | \(\ds y x x^{-1} y^{-1} x y\) | Definition of Commutator of Group Elements | ||||||||||
| \(\ds \) | \(=\) | \(\ds y y^{-1} x y\) | Group Axiom $\text G 3$: Existence of Inverse Element | |||||||||||
| \(\ds \) | \(=\) | \(\ds x y\) | Group Axiom $\text G 3$: Existence of Inverse Element |
Let $h \in H$, $k \in K$.
We have:
| \(\ds \sqbrk {h, k}\) | \(=\) | \(\ds h^{-1} k^{-1} h k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds h^{-1} \paren {k^{-1} h k}\) | Group Axiom $\text G 1$: Associativity | |||||||||||
| \(\ds \) | \(\in\) | \(\ds h^{-1} H\) | Definition of Normal Subgroup | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqbrk {h, k}\) | \(\in\) | \(\ds H\) | as $h^{-1} H = H$ |
and:
| \(\ds \sqbrk {h, k}\) | \(=\) | \(\ds h^{-1} k^{-1} h k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {h^{-1} k^{-1} h} k\) | Group Axiom $\text G 1$: Associativity | |||||||||||
| \(\ds \) | \(\in\) | \(\ds K k\) | Definition of Normal Subgroup | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqbrk {h, k}\) | \(\in\) | \(\ds K\) | as $K k = K$ |
Thus:
- $\sqbrk {h, k} \in H \cap K$
But as $H \cap K = \set e$, it follows that:
- $\sqbrk {h, k} = e$
It follows from Commutator is Identity iff Elements Commute that:
- $h k = k h$
and the result follows.
$\blacksquare$