Question

Let G be a group and let $\text{a}\in \text{G}$. Prove that $〈\text{a}〉\text{=}〈{\text{a}}^{-\mathbf{1}}〉$.

Short answer

Answer

It is proved that $〈\text{a}〉\text{=}〈{\text{a}}^{-\mathbf{1}}〉$.

Step-by-step solution

Step-by-Step Solution Step 1: Key concept

By the definition for all $\text{a}\in \text{G}$, we have that $\text{b}\in 〈\text{a}〉$if and only if there is $\text{n}\in \mathbb{Z}$such that ${\text{b=a}}^{\text{n}}$.

Prove that a=a−1

Suppose $a^n\in 〈a〉$, then $a^n={\left(a^{-1}\right)}^{-n}$, and so $a^n\in 〈a^{-1}〉$, and since aⁿ was chosen arbitrarily then, $〈a〉\subseteq 〈a^{-1}〉$.

Conversely, suppose that ${\left(a^{-1}\right)}^n\in 〈a^{-1}〉$, then ${\left(a^{-1}\right)}^n=a^{-n}so{\left(a^{-1}\right)}^n\in 〈a〉$and therefore,$〈a^{-1}〉\subseteq 〈a〉$ .

Thus, it is proved that $〈a〉=〈a^{-1}〉$.