Group Direct Product/Examples/R-0 x R

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Example of Group Direct Product

Let $G$ be the Cartesian product of $\R \setminus \set 0$ with $\R$:

$G = \set {\tuple {a, b} \in \R^2: a \ne 0}$

Let $\circ$ be a group product on $G$ defined as:

$\tuple {a_1, b_1} \circ \tuple {a_2, b_2} = \tuple {a_1 a_2, a_1 b_2 + b_1}$

Then the algebraic structure $\struct {G, \circ}$ is a group which is non-abelian.


Isomorphism to Set of Affine Mappings on Real Line under Composition

Let $S$ be the set of all real functions $f: \R \to \R$ of the form:

$\forall x \in \R: \map f x = r x + s$

where $r \in \R_{\ne 0}$ and $s \in \R$

Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.


Then $\struct {G, \circ}$ is (group) isomorphic to $\struct {S, \circ}$.


Proof

Taking the group axioms in turn:


Group Axiom $\text G 0$: Closure

By Definition of $\circ$:

$\tuple {a_1, b_1} \circ \tuple {a_2, b_2} = \tuple {a_1 a_2, a_1 b_2 + b_1}$

where $a_1, a_2 \ne 0$.

Hence:

$a_1 a_2 \ne 0$

from which:

$\tuple {a_1 a_2, a_1 b_2 + b_1} \in \paren {\R \setminus \set 0} \times \R$

Thus $\circ$ is closed in $G$.

$\Box$


Group Axiom $\text G 1$: Associativity

\(\ds \paren {\tuple {a_1, b_1} \circ \tuple {a_2, b_2} } \circ \tuple {a_3, b_3}\) \(=\) \(\ds \tuple {a_1 a_2, a_1 b_2 + b_1} \circ \tuple {a_3, b_3}\)
\(\ds \) \(=\) \(\ds \tuple {a_1 a_2 a_3, a_1 a_2 b_3 + a_1 b_2 + b_1}\)
\(\ds \tuple {a_1, b_1} \circ \paren {\tuple {a_2, b_2} \circ \tuple {a_3, b_3} }\) \(=\) \(\ds \tuple {a_1, b_1} \circ \tuple {a_2 a_3, a_2 b_3 + b_2}\)
\(\ds \) \(=\) \(\ds \tuple {a_1 a_2 a_3, a_1 a_2 b_3 + a_1 b_2 + b_1}\)

Thus $\circ$ is associative.

$\Box$


Group Axiom $\text G 2$: Existence of Identity Element

\(\ds \tuple {a_1, b_1} \circ \tuple {1, 0}\) \(=\) \(\ds \tuple {a_1 \times 1, a_1 \times 0 + b_1}\)
\(\ds \) \(=\) \(\ds \tuple {a_1, b_1}\)


\(\ds \tuple {1, 0} \circ \tuple {a_2, b_2}\) \(=\) \(\ds \tuple {1 \times a_2, 1 \times b_2 + 0}\)
\(\ds \) \(=\) \(\ds \tuple {a_2, b_2}\)

Thus $\tuple {1, 0}$ is the identity element of $\struct {G, \circ}$.

$\Box$


Group Axiom $\text G 3$: Existence of Inverse Element

We have that $\tuple {1, 0}$ is the identity element of $\struct {G, \circ}$.

We need to find $\tuple {a_2, b_2} \in G$ such that:

\(\ds a_1 a_2\) \(=\) \(\ds 1\)
\(\ds a_1 b_2 + b_1\) \(=\) \(\ds 0\)

Hence:

\(\ds a_2\) \(=\) \(\ds \dfrac 1 {a_1}\)
\(\ds b_2\) \(=\) \(\ds -\dfrac {b_1} {a_1}\)


We verify:

\(\ds \tuple {a_1, b_1} \circ \tuple {\dfrac 1 {a_1}, -\dfrac {b_1} {a_1} }\) \(=\) \(\ds \tuple {a_1 \times \dfrac 1 {a_1}, a_1 \times \paren {-\dfrac {b_1} {a_1} } + b_1}\)
\(\ds \) \(=\) \(\ds \tuple {1, 0}\)
\(\ds \tuple {\dfrac 1 {a_1}, -\dfrac {b_1} {a_1} } \circ \tuple {a_1, b_1}\) \(=\) \(\ds \tuple {\dfrac 1 {a_1} \times a_1, \dfrac 1 {a_1} \times b_1 + \paren {-\dfrac {b_1} {a_1} } }\)
\(\ds \) \(=\) \(\ds \tuple {1, 0}\)

Thus every element $\tuple {a_1, b_1}$ of $\struct {G, \circ}$ has an inverse $\tuple {\dfrac 1 {a_1}, -\dfrac {b_1} {a_1} }$.

$\Box$


All the group axioms are thus seen to be fulfilled, and so $\struct {G, \circ}$ is a group.

It remains to show that $\struct {G, \circ}$ is not abelian.

\(\ds \tuple {a_1, b_1} \circ \tuple {a_2, b_2}\) \(=\) \(\ds \tuple {a_1 a_2, a_1 b_2 + b_1}\)
\(\ds \tuple {a_2, b_2} \circ \tuple {a_1, b_1}\) \(=\) \(\ds \tuple {a_2 a_1, a_2 b_1 + b_2}\)

Selecting arbitrary values, let:

$\tuple {a_1, b_1} = \tuple {1, 2}$
$\tuple {a_2, b_2} = \tuple {3, 4}$

Then:

$a_1 b_2 + b_1 = 1 \times 4 + 2 = 6$
$a_2 b_1 + b_2 = 3 \times 2 + 4 = 10$

As it is not generally true that $a_1 b_2 + b_1$ equals $a_2 b_1 + b_2$, it follows that $\circ$ is not commutative on $G$.

$\blacksquare$


Sources

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