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I was wondering if the following proof looks okay.

Proof. Let $g$ be any element of a group $G$. Then by definition, the order of $g$ is the least positive integer $n$ such that $g^{n}=e$. We define the order of the element $g$ as $|g|=n$. Now, we wish to show that in any group, an element and its inverse have the same order.

For the inverse of the group element g,

$(g^{-1})^{n}=g^{-n}=(g^{n})^{-1}=e^{-1}=e$.

Since the order of $g^{-1}$ is the least positive integer $m$ such that $(g^{-1})^{m}=e$, it follows that $|g^{-1}|\le |g|=n$.

Likewise, $((g^{-1})^{-1})^{m})=(g^{-1})^{-m}=(g^{-m})^{-1}=((g^{m})^{-1})^{-1}=(e^{-1})^{-1}=e$

Then since $(g^{-1})^{-1}=g$, $|g|\le |g^{-1}|.$

Therefore, the order of $g$ is equal to the order of its inverse. QED

Xam
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Skm
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    Please do not as for proof-verification unless you have some concrete question about your proof. If you have a proof of a problem with no problems you're aware of, post it as a self-answered question. But first search to see if the question has been asked before, and put the solution there. In this case, that might be here. If you find that nobody has asked the question, then self-answering is the way to go. – rschwieb Sep 25 '17 at 18:39
  • I love these kinds of proofs, where we have $F(f(f(a))=a)\ge F(f(a))\ge F(a)$ therefore $F(f(a))=F(a)$. Elegant. – anonymous67 Sep 25 '17 at 18:40
  • I would like to ask at this point: the line that contains "it follows that ord(g^-1) < ord(g) = n", why does that conclusion follow?

    Bit stuck here when trying to follow this proof, otherwise it is a very nicely put together proof.

     – thesmallprint Oct 03 '17 at 16:59
  • @thesmallprint: If a group element $g$ has an order of $n$, then it follows that if $g^{k}=e$, for some $k \in \mathbb {Z^+}$, then $n$ divides $k$. So, applying it to the proof above, $(g^{-1})^{n}=e$ and $|g^{-1}|=m$. Then we have that $m$ divides $n$, so it must be that $m\leqslant n$. – Skm Oct 03 '17 at 18:59
  • Ah, I see now, thank you very much! – thesmallprint Oct 03 '17 at 20:36

2 Answers2

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It's okay for me.

Maybe it could be shortened observing it is enough to prove that $$ \forall n,\quad g^n=1\implies \bigl(g^{-1}\bigr){}^n=1. $$

Bernard
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Once you've shown that $(g^{-1})^n=e$, you could just argue that if the order of $g^{-1}$ were actually $m\in\mathbb{N}_0$, with $m<n$, then \begin{align} g^n &= e\\ (g^{-1})^mg^n&=e\\ g^{n-m}&=e \end{align} but since $n-m<n$, this is a contradiction, since $n$ was the smallest natural number such that $g^n=e$. Hence $n$ is also the smallest natural number such that $(g^{-1})^n=e$.

Stijn D'hondt
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