1. ## prove

prove that in any group, an element and its inverse have the same order.

2. Originally Posted by mandy123
prove that in any group, an element and its inverse have the same order.
let $\displaystyle a$ be in $\displaystyle G$ and let $\displaystyle n\in \mathbb{Z}^+$ be the order of $\displaystyle a$, i.e. $\displaystyle n$ be the smallest positive integer such that $\displaystyle a^n=e$

then $\displaystyle (a^{-1})^n = e$.

suppose there exists a positive integer $\displaystyle m <n$ such that $\displaystyle (a^{-1})^m= e$. then $\displaystyle \underbrace{a^{-1}a^{-1}\cdots a^{-1}}_{m \mbox{ copies}} = e$ implies that $\displaystyle \underbrace{aa\cdots a}_{m \mbox{ copies}} = a^m = e$ which is a contradiction.

hence, order of $\displaystyle a^{-1}$ is also $\displaystyle n$.