**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.2 Exercise 1.2.6**

Let $G$ be a group, and let $x,y \in G$ be elements of order 2. Prove that if $t = xy$ then $tx = xt^{-1}$ (so that if $n = |xy| < \infty$ then $x$ and $t$ satisfy the same relations in $G$ as $s$ and $r$ in $D_{2n}$).

Solution: We have $$xt^{-1} = x (xy)^{-1} = x y^{-1} x^{-1} = xyx = tx$$ since $x$ and $y$ have order 2.