Internal Group Direct Product Commutativity/Proof 2

From ProofWiki
Jump to navigation Jump to search

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:

$\forall h \in H, k \in K: h \circ k = k \circ h$


Proof

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.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Internal_Group_Direct_Product_Commutativity/Proof_2&oldid=565321"