Question

Let \( G = \mathbb{R}^* \times \mathbb{R} \), where \( \mathbb{R} \) is the set of all real numbers and \( \mathbb{R}^* = \mathbb{R} - \{0\} \). Define an operation \( \circ \) on \( G \) as \( (a,b) \circ (c,d) = (ac, bc + d) \). Then, the identity element of the group \( (G, \circ) \) is _____.

• A. \( (0,0) \)
• B. \( (0,1) \)
• C. \( (1,0) \)
• D. \( (1,1) \)

Hint:

For operations involving pairs, the identity of the first component usually matches the standard multiplicative identity (1) if the operation is product-based, and the second component often matches the additive identity (0).

Answer 1

The Correct Option is C

Concept:
In a group \( (G, \circ) \), an element \( e = (e_1, e_2) \) is called the identity element if for every element \( (a, b) \in G \): \[ (a, b) \circ (e_1, e_2) = (a, b) \quad \text{and} \quad (e_1, e_2) \circ (a, b) = (a, b) \]

Step 1: Set up the identity equation.
Let \( (e_1, e_2) \) be the identity. Applying the defined operation: \[ (a, b) \circ (e_1, e_2) = (a \cdot e_1, b \cdot e_1 + e_2) \] We want this to be equal to \( (a, b) \).

Step 2: Solve for \( e_1 \) and \( e_2 \).
Comparing components:
• \( a \cdot e_1 = a \Rightarrow e_1 = 1 \) (Since \( a \in \mathbb{R}^* \), \( a \neq 0 \))
• \( b \cdot e_1 + e_2 = b \Rightarrow b(1) + e_2 = b \Rightarrow b + e_2 = b \Rightarrow e_2 = 0 \) So, the candidate identity is \( (1, 0) \).

Step 3: Verify with left identity.
Check if \( (1, 0) \circ (a, b) = (a, b) \): \[ (1, 0) \circ (a, b) = (1 \cdot a, 0 \cdot a + b) = (a, b) \] The conditions are satisfied. Thus, \( (1, 0) \) is the unique identity element.