Question

The set $\left\{-1,0,1\right\}$ is not a multiplicative group because of the failure of

• A. Identity law
• B. Inverse law
• C. Closure law
• D. Associative law

Answer 1

The correct option is B

Inverse law

Explanation of correct option

Given, $A=\left\{-1,0,1\right\}$

A multiplicative group is a set which is associated with multiplication and obeys that inverse law, identity law, closure law and the associative law. The laws can be explained as follows:

1. Identity Law: There exists an element called identity element such that the operation between any element and the identity element results in the element itself.
2. Inverse Law: There exists a pair of elements between whom the operation results in the identity element.
3. Closure Law: All the possible results of the operation between any elements are elements of the group.
4. Associative law: The order of the operands does not affect the result.

The inverse of $1$ is $\frac{1}{1}=1\in A$
The inverse of $-1$ is $\frac{1}{-1}=-1\in A$
The inverse of $\frac{1}{0}=\infty \notin A$
The group does not satisfy the inverse law

Hence option B is option.

Explanation for incorrect option:

Option A: The identity element of multiplication is $1$ and it is an element of $A$.
Hence, option A is incorrect.

Option C: We can see that,
$$\begin{gathered} 1\times -1=-1\in A \\ 1\times 0=0\in A \\ -1\times 0=0\in A \\ 1\times 1=1\in A \\ -1\times -1=1\in A \\ 0\times 0=0\in A \end{gathered}$$

Thus, the group satisfies closure law.
Hence, option C is also incorrect.

Option D: We know that multiplication is an associative property.
$$\begin{gathered} 1\times -1=-1\times 1=-1 \\ 1\times 0=0\times 1=0 \\ -1\times 0=0\times -1=0 \end{gathered}$$

Thus, the group also satisfies associative law.
Hence, option D is also incorrect

Therefore, only option B is correct