Exercise 1.1.33 ($x^i = x^{-i}$)

Proof.

(a)

Reasoning by contradiction, assume that $x^i=x^{-i}$, where $1\le i<n$. Then $x^{2i}=e$. Since $|x|=n$, $n\mid 2i$. Here $n$ is odd, so $\gcd (n,2)=1$. This shows that $n\mid i$, where $i\ne 0$, thus $n\le i$. This is a contradiction, since $1\le i<n$.

If $n$ is odd then $x^i\ne x^{-i}$ for all $i=1,2,\dots ,n-1$.

(b)

Assume now that $n=2k$ is even. Then $$\begin{array}{llll}\hfill x^i=x^{-i}& ⟺x^{2i}=e& \hfill & \\ \hfill & ⟺|x|=n\mid 2i& \hfill & \\ \hfill & ⟺\frac{n}{2}\mid i.& \hfill & \end{array}$$

If $i=n∕2=k$, then $\frac{n}{2}\mid i$, so $x^i=x^{-I}$ by the preceding equivalence.

Conversely, if $x^i=x^{-i}$, then $i=q\frac{n}{2}$ for some integer $q$. Since $1\le i<n$, $1\le q\frac{n}{2}<n$, thus $1\le q<2$, so $q=1$, and $i=n∕2=k$.

If $n=2k$ and $1\le i<n$ then $x^i=x^{-i}$ if and only if $i=k$.