Group Direct Product/Examples/R-0 x R

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.

Proof
Taking the group axioms in turn:

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 $G$.

Thus $\circ$ is associative.

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

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:

Hence:

We verify:

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

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.

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 $G$.