Question

Let G = {(a, b) | a, b are real number, b != 0}. Define (a, b) * (c, d) = (a + bc, bd) for all (a, b), (c, d) belongs to G. Then (G, *) is a

a)Commutative group
b)non-commutative group
c)not a group
d)cyclic group.

Answer 1

1) Associativity

$$\begin{array}{l} \left( \begin{array}{c} a \\ b \end{array} \right) * \left( \begin{array}{c} c \\ d \end{array} \right) * \left( \begin{array}{c} e \\ f \end{array} \right) = \left( \begin{array}{c} a \\ b \end{array} \right) * \left( \begin{array}{c} c + d e \\ d f \end{array} \right) = \left( \begin{array}{c} a + b c + b d e \\ b d f \end{array} \right) \\ \left(\left( \begin{array}{l} a \\ b \end{array} \right) * \left( \begin{array}{l} c \\ d \end{array} \right)\right) * \left( \begin{array}{l} e \\ f \end{array} \right) = \left( \begin{array}{l} a + b c \\ b d \end{array} \right) * \left( \begin{array}{l} e \\ f \end{array} \right) = \left( \begin{array}{l} a + b c + b d e \\ b d f \end{array} \right) \\ \left( \begin{array}{c} a \\ b \end{array} \right) * \left( \begin{array}{c} c \\ d \end{array} \right) * \left( \begin{array}{c} e \\ f \end{array} \right) = \left( \begin{array}{c} a \\ b \end{array} \right) * \left( \begin{array}{c} c \\ d \end{array} \right) * \left( \begin{array}{c} e \\ f \end{array} \right) \end{array}$$

2) Identity

$$\left( \begin{array}{c} a \\ b \end{array} \right) * \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \left( \begin{array}{c} a \\ b \end{array} \right) = \left( \begin{array}{c} 0 \\ 1 \end{array} \right) * \left( \begin{array}{c} a \\ b \end{array} \right)$$

3) Inverse:

$$\left( \begin{array}{c} a \\ b \end{array} \right) * \left( \begin{array}{c} - a / b \\ 1 / b \end{array} \right) = \left( \begin{array}{c} - a / b \\ 1 / b \end{array} \right) * \left( \begin{array}{c} a \\ b \end{array} \right) = \left( \begin{array}{c} 0 \\ 1 \end{array} \right)$$

4) Commutativity:

$$\begin{array}{l} \left( \begin{array}{c} a \\ b \end{array} \right) * \left( \begin{array}{c} c \\ d \end{array} \right) = \left( \begin{array}{c} a + b c \\ b d \end{array} \right) \\ \left( \begin{array}{l} c \\ d \end{array} \right) * \left( \begin{array}{l} a \\ b \end{array} \right) = \left( \begin{array}{l} c + a d \\ d b \end{array} \right) \\ a + b c \neq c + a d \end{array}$$

So it is non-commutative group.