Question

Let G be an abelian group and n a fixed positive integer. Prove that$\text{H=}\left\{{\text{a}}^{\text{n}}\text{|a}\in \text{G}\right\}$ is a subgroup of G.

Short answer

Answer

It is proved that$\text{H=}\left\{{\text{a}}^{\text{n}}\text{|a}\in \text{G}\right\}$ is a subgroup of G.

Step-by-step solution

Step by Step Solution Step 1: Required Theorem

A non-empty subset H of group G is a subgroup of G provided that,

$$\begin{gathered} \text{(i)Ifa,b}\in \text{H,thenab}\in \text{H;and} \\ \left(ii\right)If\text{a}\in {\text{Hthena}}^{-\mathbf{1}}\in \text{H} \end{gathered}$$

Prove that H=an|a∈G is a subgroup of G

$$\begin{gathered} Letx=a^{n}andy=a^{n}betworandomelementsofH.AsHisanabeliangroup,wecaninferthat, \\ x\cdot y=a^n\cdot b^n={\left(ab\right)}^n\in H \\ Letx=a^{n}beanarbitraryelementofH. \\ Since{x}^{-1}={\left(a^{-1}\right)}^n,so{x}^{-1}={\left(a^n\right)}^{-1} \\ So,x^{-1}\in H. \end{gathered}$$

Concluding the results

On comparing the above two results with theorem 7.11, we can conclude that$H=\left\{a^n|a\in G\right\}$is a subgroup of G.