Prove that x and x^-1 have same order

[Mr Davis 97]

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]

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 $a=b\,?$