| 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 |
| 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 |
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
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\begin{vmatrix} \text {x + 4} & \text{2x} & \text{2x} \\ \text{ 2x} & \text{x + 4} & \text{2x} \\ \text{2x} & \text{2x} & \text{x + 4} \end{vmatrix} = ( 5\text{x} + 4) (4 -\text{x}) = 1. $