<s>[INST] Given $x^9 = e$ and $x^{11} = e$ prove $x = e$. 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, $x = e$.
 [/INST] Yes, it is that simple. It can be done even shorter, because after showing $x^2=e$, you can go straight to
$$e=x^9=x(x^2)^4=xe^4=x$$and you're done.
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