Proof verification of an elementary exercise in abstract algebra Problem: let $x\in G$.
$|x|=n$, the order of $x$, is an odd number.
Now prove that $x^i\neq x^{-i}$ for all $i=1,2,\ldots, n-1$.
My attempt: Let's assume $|x|=n=2k+1$ for some integer $k>0$. Suppose otherwise, that $x^i=x^{-i}$ for some $i=1,2,\ldots,$ or $n-1$. Then
$$x^{2k+1}=1$$$$x^{2k+1-i}=x^{-i}=x^i$$$$x^{2k+1-2i}=1$$i.e. $x^{n-2i}=1$. Since $i$ can take a range from $1$ to $n-1$, we have $n-2i\in[-n+2,n-2]$, an odd number, and so it is non-zero. If $n-2i>0$, then it contradicts the minimality of $n$ being the smallest positive power of $x$ eq. to $1$. Otherwise, $x^{-(n-2i)}=1$, a power of $x$ also $<n$. We reach the same contradiction as before. $\blacksquare$
Can someone help me by verifying the above proof?

That works. A slightly faster (but essentially the same) way to see it is as follows. Suppose $|x| = n$ odd, and suppose $x^i = x^{-i}$ for some $i \in \{1,\dots,n-1\}$. Then $x^{2i} = 1$. Now since $|x| = n$ we must have $2i > n$ (since $n$ is odd, we can't have equality). But in addition, $2i < 2n$ by assumption, so $n < 2i < 2n$, so $0 < 2i-n < n$, and thus we get $x^{2i-n} = x^{2i}(x^n)^{-1} = 1$, which is a contradiction.
