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: Group Definition
A group \((G, *)\) is defined by 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: There exists \(e \in G\) such that for all \(a \in G\), \(a*e = e*a = a\).
4. Inverse: For each \(a \in G\), there exists \(a^{-1} \in G\) such that \(a*a^{-1} = a^{-1}*a = e\).

Step 2: Analysis of Options
Let's examine each option:
(A) \( (\mathbb{N}, *)\), where \(a*b = a\):

• Associativity: \((a*b)*c = a*c = a\). And \(a*(b*c) = a*b = a\). Holds.
• Identity: Need \(e \in \mathbb{N}\) where \(a*e=a\) and \(e*a=a\). \(a*e=a\) is always true. \(e*a = e = a\). No unique identity. 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 \(a*(b*c) = a*(b-c) = a-(b-c) = a-b+c\). Not associative. Not a group.

(C) \( (\mathbb{R}, *)\), where \(a*b = a+b+1\):

• Closure: If \(a,b \in \mathbb{R}\), then \(a+b+1 \in \mathbb{R}\). Holds.
• Associativity: \((a*b)*c = (a+b+1)*c = (a+b+1)+c+1 = a+b+c+2\). And \(a*(b*c) = a*(b+c+1) = a+(b+c+1)+1 = a+b+c+2\). Holds.
• Identity: Need \(a*e = a\), so \(a+e+1=a\), giving \(e=-1\). Check: \(-1+a+1=a\). Identity is \(e=-1\), which is in \(\mathbb{R}\).
• Inverse: Need \(a*a^{-1}=e\), so \(a+a^{-1}+1 = -1\), giving \(a^{-1} = -a-2\). Since \(a \in \mathbb{R}\), \(-a-2\) is also in \(\mathbb{R}\). An inverse exists.
• All axioms are satisfied. This is a group.

(D) \( (\mathbb{R}, *)\), where \(a*b = |a|b\):

• Identity: Need \(a*e=a\) and \(e*a=a\). \(|a|e=a \implies e=a/|a|\). No unique identity. Not a group.

Step 3: Conclusion
The structure \( (\mathbb{R}, *)\) with \(a*b = a+b+1\) is the only group.