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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