Question:
Consider the binary operation \( \circ \) defined on the set \( S = \{a, b, c, d\} \) by the following table:
\[ \begin{array}{c|cccc} \circ & a & b & c & d \\ \hline a & a & a & a & a \\ b & a & b & c & d \\ c & a & c & d & b \\ d & a & d & b & c \\ \end{array} \]
Solution & Observations:
1. Closure:
All elements in the table belong to \( S \). Hence, the operation is closed.
2. Identity Element:
Check if any element satisfies:
\[ e \circ x = x \quad \text{and} \quad x \circ e = x \]
From the table:
- Row of b: \( b \circ a = a,\; b \circ b = b,\; b \circ c = c,\; b \circ d = d \)
- Column of b: \( a \circ b = a,\; b \circ b = b,\; c \circ b = c,\; d \circ b = d \)
Thus, b acts as identity element.
3. Inverses:
- \( a \circ b = a \Rightarrow a^{-1} = b \)
- \( b \circ b = b \Rightarrow b^{-1} = b \)
- \( c \circ d = b \Rightarrow c^{-1} = d \)
- \( d \circ c = b \Rightarrow d^{-1} = c \)
4. Commutativity:
Check symmetry of table:
Since \( c \circ d = b \) and \( d \circ c = b \), and similar symmetry holds → operation is commutative.
5. Associativity:
From structure and consistency, operation behaves associatively.
Final Conclusion:
The given binary operation is a commutative group with identity element \( b \). Each element has an inverse, and the operation satisfies closure and associativity.