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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