Question
Consider the binary operation * and 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

Answer

Commutativity:The table is symmetrical about the leading element. It means that o is commutative on S.
$a o (b o c) = a o c$
$= a$
$(a o b) o c = a o c$
$= a$
thus,
a o (b o c) = (a o b) o $\text{c }\forall\text{ a, b, c}\in\text{S}$
Associativity:
Therefore, 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.
Implies that x o b = b o x
$=\text{x, }\forall\text{ x}\in\text{S}$
Therefore,
b is the identity element.
Finding inverse elements:
In the first row, we don't have b,
i.e. there does not exist an element x such that a o x = x o a = b.
Therefore,
$a^{-1}$​​​​​​​ does not exists.
b o b = b
Implies that $b^{-1}= b$
c o d = b
Implies that $c^{-1} = d$
d o c = b
Implies that $d^{-1} = c$

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