Question

Consider the binary operation * and o defined by the following tables on set S = {a, b, c, d}.
(i)

*	a	b	c	d
a	a	b	c	d
b	b	a	d	c
c	c	d	a	b
d	d	c	b	a

(ii)

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 both the binary operations are commutative and associatve. Write down the identities and list the inverse of elements.

Answer 1

(i) Commutativity:
The table is symmetrical about the leading element. It means * is commutative on S.

Associativity:
$$\begin{gathered} a*\left(b*c\right)=a*d \\ =d \\ \left(a*b\right)*c=b*c \\ =d \\ \mathrm{Therefore}\text{,} \\ a*\left(b*c\right)=\left(a*b\right)*c\forall a,b,c\in S \end{gathered}$$

So, * is associative on S.

Finding identity element:
We observe that the first 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 a.

$\Rightarrow x*a=a*x=x,\forall x\in S$

So, a is the identity element.

Finding inverse elements:
$$\begin{gathered} a*a=a \\ \Rightarrow a^{-1}=a \\ b*b=a \\ \Rightarrow b^{-1}=b \\ c*c=a \\ \Rightarrow c^{-1}=c \\ d*d=a \\ \Rightarrow d^{-1}=d \end{gathered}$$

(ii) Commutativity:
The table is symmetrical about the leading element. It means that o is commutative on S.
Associativity:
$$\begin{gathered} ao\left(boc\right)=aoc \\ =a \\ \left(aob\right)oc=aoc \\ =a \\ \text{Thus,} \\ ao\left(boc\right)=\left(aob\right)oc\forall a,b,c\in S \end{gathered}$$

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.

$$\begin{gathered} \Rightarrow xob=box \\ =x,\forall x\in S \end{gathered}$$

So, b is the identity element.

Finding inverse elements:
$$\begin{gathered} \text{In the first row, we don't have}b,\text{i.e. there does not exist an element}\mathit{\text{x}}\text{such that}aox=xoa=b. \\ \text{So,}{a}^{-1}\text{does not exist.} \\ bob=b \\ \Rightarrow b^{-1}=b \\ cod=b \\ \Rightarrow c^{-1}=d \\ doc=b \\ \Rightarrow d^{-1}=c \end{gathered}$$