Question

Find the multiplicative inverse of each nonzero element in

(a)${\mathbf{\mathbb{Z}}}_{\mathbf{3}}$.

Short answer

It is a group of order n. It is abelian as well as cyclic in nature. If 1 is the identity element then it satisfies the relation ${\mathit{a}}^{\mathbf{n}}\mathbf{=}\mathbf{1}$, it represents cyclicity and the multiplication table is given by:

Step-by-step solution

Step: 1: Definition of ℤ n:

It is a group of order n. It is abelian as well as cyclic in nature. If 1 is the identity element then it satisfies the relation ${\mathit{a}}^{\mathbf{n}}\mathbf{=}\mathbf{1}$, it represents cyclicity and the multiplication table is given by:

Step: 2: Definition of ℤ3:

It is a group of order 3. It is abelian as well as cyclic in nature. If 1 is the identity element then it satisfies the relation ${\mathit{a}}^{\mathbf{3}}\mathbf{=}\mathbf{1}$, it represents cyclicity and the multiplication table is given by:

Step: 3: Inverse of an element:

Inverse of an element $\mathit{a}$ in a group G is defined as an element$\mathit{a}\mathbf{\text{'}}$such that,

$\mathit{a}\mathit{a}\mathbf{\text{'}}\mathbf{=}\mathit{a}\mathbf{\text{'}}\mathit{a}\mathbf{=}\mathit{e}$where $\mathit{e}$ is the identity element in G.

Step: 4: Complete the table

By the table we can see the non-zero elements are 1 and 2.

Their inverses are given by: