Question

Abstract Algebra

Let X be a set and let P(X) be the power set of X.

a) Does P(X) with the binary operation A *B=ANB form a group? Justify your answer.

b) Does P(X) with the binary operation A *B=AUB form a group? Justify your answer.

Answer 1

1.No, because inverse does not exist.

2.No, because inverse does not exist.

Explanation:

We are given that X be a set and let P(X) be the power set of X.

a. We have to tell P(X) with binary operation

A*B=
`A\cap B` form a group.

Suppose, x={1,2}

P(X)={
`\phi`,{1},{2},{1,2}}

1.Closure property:
`A\cap B\in P(X)`

{1}
`\cap` {2}=
`\phi \in P(X)`

It is satisfied for all
`A,B\in P(X)`

2.Associative property:
`(A\cap B)\cap C=A\cap (B\cap C)`

If A={1},B={2},C={1,2}

`A\cap(B\cap C)`={1}
`\cap`({2}
`\cap`{1,2})={1}
`\cap` {2}=
`\phi`

`(A\cap B)\cap C`=({1}
`\cap`{2})
`\cap`{1,2}=
`\phi\cap`{1,2}=
`\phi`

Hence, P(X) satisfied the associative property.

3.Identity :
`A\cap B=A` Where B is identity element of P(X)

`A\cap X=A`

It is satisfied for every element A in P(X).

Hence, X is identity element in P(X)

4.Inverse :
`A\cap B=X` Where B is an inverse element of A in P(x)

It can not be possible for every element that satisfied
`A\cap B=X`

Hence, inverse does not exist.

Therefore, P(X) is not a group w.r.t to given binary operation.

2.We have to tell P(X) with the binary operation

A*B=
`A\cup B` form a group

Similarly,

For set X={1,2}

P(X)={
`\phi`,{1},{2},{1,2}}

1.Closure property:If A and B are belongs to P(X) then
`A\cup B\in P(X)` for all A and B belongs to P(X).

2.Associative property:
`A\cup (B\cup C)=(A\cup B)\cup C`

If A={1},B={2},C{1,2}

`A\cup B`={1}
`\cup`{2}={1,2}

`(A\cup B)\cup C`={1,2}
`\cup`{1,2}={1,2}

`B\cup C`={2}
`\cup`{1,2}={1,2}

`A\cup (B\cup C)`={1}
`\cup`{1,2}={1,2}

Hence, P(X) satisfied the associative property.

3.Identity :
`A\cup B=A` Where B is identity element of P(X)

Only
`\phi` is that element for every A in P(X) that satisfied
`A\cup B=A`

Hence,
`\phi` is identity element of P(X) w.r.t union.

4.Inverse element :

`A\cup B=\phi` where B is an inverse element of A in P(X)

It is not possible for every element that satisfied the property.

Hence, inverse does not exist for each element in P(X).

Therefore, P(X) is not a group w.r.t binary operation.