Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.16
Let $G$ be a group and let $x \in G$. Prove that $x^2 = 1$ if and only if $|x|$ is either 1 or 2.
Solution:
($\Rightarrow$) Suppose $x^2 = 1$. Then we have $0 < |x| \leq 2$, i.e., $|x| $is either 1 or 2.
($\Leftarrow$) If $|x| = 1$, then we have $x = 1$ so that $x^2 = 1$. If $|x| = 2$ then $x^2 = 1$ by definition. So if $|x|$ is 1 or 2, we have $x^2 = 1$.
