Question

Let $\mathit{G}\mathbf{=}\mathbf{\{}{\mathbf{a}}_{\mathbf{1}}\mathbf{,}{\mathbf{a}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{a}}_{\mathbf{n}}\mathbf{\}}$be a finite abelian group of order $\mathit{n}$. Let role="math" localid="1653675055865" $\mathit{x}\mathbf{=}{\mathit{a}}_{\mathbf{1}}{\mathit{a}}_{\mathbf{2}}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{a}}_{\mathbf{n}}$. Prove thatrole="math" localid="1653675071281" ${\mathit{x}}^{\mathbf{2}}\mathbf{=}\mathit{e}$.

Short answer

It is proved that $x^2=e$.

Step-by-step solution

Determine given function

Consider the given function as $G=\left\{a_1,a_2,...,a_n\right\}$ is a finite abelian group and order of G is $n$.

Determine x2=e

Let’s consider that, is the bijection.

So that, this exists by the axioms that the elements of the group have inverse and inverse groups are unique.

Using the commutative property in as follows

$$\begin{gathered} x^2={\left(a_{1........}{a}_n\right)}^2 \\ =a_{1.....}{a}_n{a}_{1....}{a}_n \\ =a_1{a}_{\sigma \left(1\right)....}{a}_n{a}_{\sigma \left(n\right)} \\ =a_1{a_1}^{-1}.....a_n{a_n}^{-1} \end{gathered}$$

This implies, $x^2=e$.

Hence, $x^2=e$.