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

Question

Elberte · Accepted Answer

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