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

Prove that in any group, an element and its inverse have the same order.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Elberte

Proof:
Let x be a element in a group and ${x}^{-1}$ be its inverse.
Assume $o\left(x\right)=m\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}o\left({x}^{-1}\right)=n$
It is know that if ${a}^{p}=e$, then $o\left(a\right)=p$
consider, $o\left(x\right)=m$ and simplify as shown below
${x}^{m}=e$
${\left({x}^{m}\right)}^{-1}={e}^{-1}$
${x}^{-m}=e$
$o\left({x}^{-1}\right)\le m$
$n\le m$ (1)
Now consider $o\left({x}^{-1}\right)=n$ and simplify.
${\left({x}^{-1}\right)}^{n}=e$
${x}^{-n}=e$
${\left({x}^{-n}\right)}^{-1}={e}^{-1}$
${x}^{n}=e$
$o\left(x\right)\le n$
$m\le n$ (2)
From equation (1) and (2), $m=n$
Hence proved.

###### Not exactly what you’re looking for?
Jeffrey Jordon

Step 1

let G be a group and a an element of G with order n (${a}^{n}=e$),so

$\left({a}^{-1}{\right)}^{n}={a}^{-n}=\left({a}^{n}{\right)}^{-1}={e}^{-1}=e$

therefore

$order\left({a}^{-1}\right)\le order\left(a\right)$

Step 2

by the same argument we can say:

$order\left(a\right)=order\left(\left({a}^{-1}{\right)}^{-1}\right)\le order\left({a}^{-1}\right)$  we conclude that :

$order\left(a\right)\le order\left({a}^{-1}\right)$

Finally we conclude that: $order\left(a\right)=order\left({a}^{-1}\right)$