Prove the following: The order of $a^{-1}$ is the same as the order of $a$.

Would this be the correct proof?

$a^n=e$ then

$e=(a^{-1})^n$ and let $m=$ord$(a^{-1})$. Then $m \le n$.

and $(a^{-1})^m = e$ then $e=a^m$. It must be that $n \le m$. In conclusion $m=n$.

  • $\begingroup$ This is correct. $\endgroup$ – alex Mar 4 '14 at 21:47
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    $\begingroup$ I feel like you could add a bit of extra detail explaining why $a^n=e$ implies $(a^{-1})^n=e$. $\endgroup$ – Joshua Pepper Mar 4 '14 at 21:48
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    $\begingroup$ What happened to $b$ and $c$? $\endgroup$ – alex Mar 4 '14 at 21:49
  • $\begingroup$ The tag "order theory" has nothing to do with the order of an element in a group. $\endgroup$ – amWhy Mar 4 '14 at 22:00

Let $d$ be the order of $a^{-1}$.

First, $a$ and $a^{-1}$ commute. So, as $aa^{-1} = e$: $$ e = (aa^{-1})^n = a^n (a^{-1})^n $$ This proves that $d|n$.

As $(a^{-1})^{-1} = a$, by symetry we get $n|d$.

Conclusion: $d=n$.


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

Here I have used some basic group operation identities.

  • $\begingroup$ You only showed that the order of $a^{-1}$ divides $n$. $\endgroup$ – Joshua Pepper Mar 4 '14 at 22:01

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