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Question
Consider the binary operation * defined by the following tables on set S = {a, b, c, d}.
| * | a | b | c | d |
| a | a | b | c | d |
| b | b | a | d | c |
| c | c | d | a | b |
| d | d | c | b | a |
Show that the binary operation is commutative and associative. Write down the identities and list the inverse of elements.
Solution
Commutativity:
The table is symmetrical about the leading element. It means * is commutative on S.
Associativity:
\[a * \left( b * c \right) = a * d\]
\[ = d\]
\[\left( a * b \right) * c = b * c\]
\[ = d\]
\[\text{Therefore},\]
\[a * \left( b * c \right) = \left( a * b \right) * c \forall a, b, c \in S\]
So, * is associative on S.
Finding identity element :-
We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at a.
\[\Rightarrow x * a = a * x = x, \forall x \in S\]
So, a is the identity element.
Finding inverse elements :-
\[a * a = a\]
\[ \Rightarrow a^{- 1} = a\]
\[b * b = a\]
\[ \Rightarrow b^{- 1} = b\]
\[c * c = a\]
\[ \Rightarrow c^{- 1} = c\]
\[d * d = a\]
\[ \Rightarrow d^{- 1} = d\]
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