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• September 26th 2008, 08:06 AM
mandy123
prove
prove that in any group, an element and its inverse have the same order.(Headbang)
• September 26th 2008, 08:41 AM
kalagota
Quote:

Originally Posted by mandy123
prove that in any group, an element and its inverse have the same order.(Headbang)

let $a$ be in $G$ and let $n\in \mathbb{Z}^+$ be the order of $a$, i.e. $n$ be the smallest positive integer such that $a^n=e$

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

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

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