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Question
Consider the binary operation 'o' defined by the following tables on set S = {a, b, c, d}.
| o | a | b | c | d |
| a | a | a | a | a |
| b | a | b | c | d |
| c | a | c | d | b |
| d | a | d | b | c |
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 that o is commutative on S.
Associativity:
\[a o \left( b o c \right) = a o c\]
\[ = a\]
\[\left( a o b \right) o c = a o c\]
\[ = a\]
\[\text{Thus},\]
\[a o \left( b o c \right) = \left( a o b \right) o c \forall a, b, c \in S\]
So, o is associative on S.
Finding identity element :-
We observe that the second 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 b.
\[\Rightarrow x o b = b o x\]
\[ = x, \forall x \in S\]
So, b is the identity element.
Finding inverse elements :-
\[\text{In the first row, we don't haveb, i.e. there does not exist an elementxsuch thata} o x = x o a = b . \]
\[So, a^{- 1} \text{does not exist}.\]
\[b o b = b\]
\[ \Rightarrow b^{- 1} = b\]
\[c o d = b\]
\[ \Rightarrow c^{- 1} = d\]
\[d o c = b\]
\[ \Rightarrow d^{- 1} = c\]
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