# Prove that x and x^-1 have same order

Let ##|x| = n## and ##|x^{-1}| = m##. So ##x^n = 1## and ##(x^{-1})^{m} = 1##. Both of these imply that ##x^m = 1## and ##(x^{-1})^n = 1##. Hence, ##n = am## and ##m = bn## for some ##a,b \in \mathbb{N}##. If we divide both equations we find that ##a = b##. Therefore ##m/n = n/m \implies n^2 = m^2 \implies n = m##. Hence, ##x## and its inverse have the same order.

Is this a correct proof?

fresh_42
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Let ##|x| = n## and ##|x^{-1}| = m##. So ##x^n = 1## and ##(x^{-1})^{m} = 1##. Both of these imply ...
Yes, but how?
... that ##x^m = 1## and ##(x^{-1})^n = 1##. Hence, ##n = am## and ##m = bn## for some ##a,b \in \mathbb{N}##. If we divide both equations we find that ##a = b##.
How? I get ##n=am=a(bn)=(ab)n## and thus ##ab=1## resp. ##a=b=1## and so ##n=m##.
Therefore ##m/n = n/m \implies n^2 = m^2 \implies n = m##. Hence, ##x## and its inverse have the same order.

Is this a correct proof?
I could follow the first missing step in mind, but not the second. How did you get ##a=b\,? ##

Mr Davis 97