Full Problem: Prove that for any element $x$ in a group $G$ that satisfies $$x^9 = e \\ x^{11} = e,$$ where $e$ is the identity element, that $x$ itself must be $e$.
Is this as simple as showing that
- $x^{11} = x^{9} \cdot x^{2} = e \cdot x^2 \Rightarrow x^2 = e$
- $x^{9} = x^{2} \cdot x^{7} = e \cdot x^7 \Rightarrow x^7 = e$
- $x^{7} = x^{2} \cdot x^{5} = e \cdot x^5 \Rightarrow x^5 = e$
- $x^{5} = x^{2} \cdot x^{3} = e \cdot x^3 \Rightarrow x^3 = e$
- $x^{3} = x^{2} \cdot x = e \cdot x \Rightarrow x = e$
Therefore, $x = e$.