# 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.
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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.

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

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