Question:
Consider the binary operation \( * \) defined on the set \( S = \{a, b, c, d\} \) by the following table:
\[ \begin{array}{c|cccc} * & a & b & c & d \\ \hline a & a & b & c & d \\ b & b & a & d & c \\ c & c & d & a & b \\ d & d & c & b & a \\ \end{array} \]
Solution & Observations:
1. Closure:
All elements of the table belong to \( S \). Hence, the operation is closed.
2. Identity Element:
Observe row and column of \( a \):
\[ a * x = x \quad \text{and} \quad x * a = x \]
So, a is the identity element.
3. Inverses:
- \( a * a = a \Rightarrow a^{-1} = a \)
- \( b * b = a \Rightarrow b^{-1} = b \)
- \( c * c = a \Rightarrow c^{-1} = c \)
- \( d * d = a \Rightarrow d^{-1} = d \)
Each element is its own inverse.
4. Commutativity:
The table is symmetric about the diagonal ⇒ operation is commutative.
5. Associativity:
This structure behaves like a known group (Klein 4-group), hence associative.
Final Conclusion:
The given structure is a commutative group with identity element \( a \), where every element is self-inverse.