TITLE: Given $x^9 = e$ and $x^{11} = e$ prove $x = e$.
QUESTION [7 upvotes]: 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, $x = e$.

REPLY [5 votes]: It can be as simple as what you proved, but it can be shorter. Actually you can prove the following result:

If in a group, $x^m=e$ and $x^n=e$ for coprime $m$ and $n$, then $x=e$.

Indeed we have a Bézout's relation: $\; um+vn=1$, so
$$x=x^{um+vn}=(x^m)^u(x^n)^v=e^ue^v=e.$$