Question

Which one of the following mathematical structure forms a group?

• A. \( (\mathbb{N}, *)\), where \(a*b = a\) for all \(a,b \in \mathbb{N}\)
• B. \( (\mathbb{Z}, *)\), where \(a*b = a-b\), for all \(a,b \in \mathbb{Z}\)
• C. \( (\mathbb{R}, *)\), where \(a*b = a+b+1\), for all \(a,b \in \mathbb{R}\)
• D. \( (\mathbb{R}, *)\), where \(a*b = |a|b\), for all \(a,b \in \mathbb{R}\)

Hint:

When checking for group structures, associativity and the existence of a unique identity are common points of failure. For binary operations involving subtraction or division, always check associativity first as it often fails. For operations defined piecewise or with absolute values, check the identity and inverse properties carefully.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A mathematical structure \((G, *)\) forms a group if it satisfies four axioms: 1. Closure: For all \(a, b \in G\), \(a*b \in G\). 2. Associativity: For all \(a, b, c \in G\), \((a*b)*c = a*(b*c)\). 3. Identity Element: There exists an element \(e \in G\) such that for all \(a \in G\), \(a*e = e*a = a\). 4. Inverse Element: For each \(a \in G\), there exists an element \(a^{-1} \in G\) such that \(a*a^{-1} = a^{-1}*a = e\).

Step 2: Detailed Explanation:
Let's check each option:
(A) \( (\mathbb{N}, *)\), where \(a*b = a\): - Associativity: \((a*b)*c = a*c = a\). And \(a*(b*c) = a*b = a\). It is associative. - Identity: We need an element \(e \in \mathbb{N}\) such that \(a*e=a\) and \(e*a=a\). The first part, \(a*e=a\), is always true. The second part requires \(e*a = e = a\). This must hold for all \(a\), but \(e\) cannot be equal to all \(a\). No unique identity element exists. Not a group.
(B) \( (\mathbb{Z}, *)\), where \(a*b = a-b\): - Associativity: \((a*b)*c = (a-b)*c = (a-b)-c = a-b-c\). And \(a*(b*c) = a*(b-c) = a-(b-c) = a-b+c\). Since \(a-b-c \neq a-b+c\) in general, it is not associative. Not a group.
(C) \( (\mathbb{R}, *)\), where \(a*b = a+b+1\): - Closure: If \(a,b \in \mathbb{R}\), then \(a+b+1 \in \mathbb{R}\). Closure holds. - Associativity: \((a*b)*c = (a+b+1)*c = (a+b+1)+c+1 = a+b+c+2\). And \(a*(b*c) = a*(b+c+1) = a+(b+c+1)+1 = a+b+c+2\). It is associative. - Identity: We need \(a*e = a\), so \(a+e+1=a\), which gives \(e=-1\). Let's check: \(e*a = -1+a+1=a\). The identity element is \(e=-1\), which is in \(\mathbb{R}\). - Inverse: We need \(a*a^{-1}=e\), so \(a+a^{-1}+1 = -1\), which gives \(a^{-1} = -a-2\). Since \(a \in \mathbb{R}\), \(-a-2\) is also in \(\mathbb{R}\). An inverse exists for every element. All axioms are satisfied. This is a group.
(D) \( (\mathbb{R}, *)\), where \(a*b = |a|b\): - Identity: We need \(a*e=a\) and \(e*a=a\). The first part: \(|a|e=a \implies e=a/|a|\). This identity element depends on \(a\), so there is no unique identity element for the whole set. Not a group.
Step 3: Final Answer:
The structure \( (\mathbb{R}, *)\) with \(a*b = a+b+1\) is the only one that forms a group.