Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.32
Let $G$ be a group and $x \in G$ an element of finite order, say $|x| = n$.
(1) Prove that if $n = 2k+1$ is odd and $1 \leq i < n$, then $x^i \neq x^{-i}$.
(2) Prove that if $n = 2k$ is even and $1 \leq i < n$, then $x^i = x^{-i}$ if and only if $i = k$.
Solution: We begin with a lemma.
Lemma 1. Let $n,k \in \mathbb{Z}^+$. If $2 \leq k < n$ and $n|k$, then $n = k$.
Proof: We have $k = nq$ for some $q \in \mathbb{N}$. If $q \geq 2$, then since $2k \leq 2n$, we have $k < 2k \leq 2n \leq k$, a contradiction. Thus $q = 1$ and we have $k = n$. $\blacksquare$
Now for the main results.
If $x^i = x^{-i}$ for some $1 \leq i < n$, we have $x^{2i} = 1$ and thus $n|2i$ by Lemma 1 of Exercise 1.1.30. Since $n$ is odd, we have $n|i$. By the lemma, $i = n$, a contradiction.
($\Leftarrow$) Clearly, if $i = k$ then $x^{2i} = 1$, so that $x^i = x^{-i}$.
($\Rightarrow$) Suppose $x^i = x^{-i}$ for some $1 \leq i < n$. Then $x^{2i} = 1$. By the Division Algorithm we have $2i = qn + r$ for some integers $q$ and $r$ with $0 \leq r 0$, we have a contradiction, so that $r = 0$. Hence $2i = qn$. Now we must have $q > 0$, and since $0 < 2i < 2n$ we have $0 < qn < 2n$ and hence $q=1$. So $2i = 2k$ and thus $i = k$.
