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Let $G$ be a group, and $x\in G$ such that $|x| = n\in\mathbb N$. Prove that $n$ is odd implies $x^i\ne x^{-i}$ for all $i=1,2,\ldots,n-1$.

PROOF: Give $x\in G$ such that $|x| = n\in\mathbb N$, if $x^i=x^{-i}$ for some $i\in\{1,2,\ldots,n-1\}$, then $$x^i=x^{-i} \iff x^{2i} = e.$$ If $2i < n$, then we have found another positive integer such that $x^{2i} = e$. But by definition, $n$ is the least positive integer such that $x^n = e$. Thus we have a contradiction. Now, if $n < 2i$, here's where I have some trouble. If $n<2i$, then $n + k = 2i$ for some integer $k$. I'm not really sure how we'd derive a contradiction here. I know that $$n=2i - k \le 2(n-1) - k = 2n-2-k < 4i - 2-k,$$ but this is as far as I got. An easy contradiction would be to show something like $n < n-2$ (for example), but it is not completely evident where I might need to go from here to derive a contradiction. Can anyone give a slight hint?

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    Do you know Lagrange's Theorem? – 2017-01-31
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    If $n \lt 2i$ then $n$ divides $2i$ since $n$ is the order. – 2017-01-31
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    If $\exists i$ s.t. $x^i=x^{-i}$, then consider $\left\lbrace e,x^i \right\rbrace$ and try to use Lagrange's theorem. – 2017-01-31
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    @ASKASK, I personally do because I've taken a class on Abstract Algebra before. However, I have encountered this kind of problem in a graduate leveling course where we go back and revisit everything from the ground up. We haven't covered Lagrange's Theorem yet. Hence technically it would be inappropriate to use Lagrange's Theorem in this case. – 2017-01-31
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    @VikrantDesai, I understand why this is because of the Division Algorithm. But I felt like it was slightly clouding the proof because we haven't proved this result yet and hence I needed to give it as a Lemma before beginning the proof. So I was kind of hoping for a more *elementary* argument by showing it can't be less than or greater than which $n=2i$ would show that $n$ is even, completing the proof. – 2017-01-31

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If $n \lt 2i$, then there exists a $k$ such that $n+k=2i$.

Then $x^{n+k}=x^nx^k=x^{2i}=e \Rightarrow x^k=e. \; (\because x^n=e).$

Now $n+k=2i \Rightarrow n=2i-k \lt 2n-k \; \; (\because i \lt n).$

This implies $k \lt n$. But we got $x^k=e$ above. Hence this is a contradiction as $n$ is the order of $x$.

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The subgroup $H=\{1, x, x^2, \cdots, x^{n-1}\}$ of $G$ has odd cardinality $n$. If there is an $1 \leq i \lt n$ with $x^i=x^{-i}$, then $x^{2i}=1$, hence $x^i$ has an order that divides $2$. But by Lagrange's Theorem, its order also divides $|H|=n$. Since $n$ is odd, we conclude that $x^i=1$, contradicting $o(x)=n$.