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, $x = e$.
 A: $$e=(x^{11})^5=x^{55}=x^{54}\cdot x=(x^9)^{6}\cdot x=e\cdot x=x$$
As you can see, this thus follows because there is integer solutions to $11x-9y=1$, which is true because $11$ and $9$ are relatively prime.
Your approach is doing much the same, using a slow form of the Euclidean algorithm to show that $11$ and $9$ are relatively prime:
$$11=9\cdot 1 + 2\\
9=2\cdot 1 + 7\\
7=2\cdot 1 + 5\\
5=2\cdot 1 + 3\\
3=2\cdot 1 + 1$$
You could have skipped a lot of steps by doing the equivalent of $9=2\cdot 4 + 1$, as other answers have suggested.
A: 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.
A: 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
$$x=x^{um+vn}=(x^m)^u(x^n)^v=e^ue^v=e.$$
A: Your reasoning is correct but here is a more direct argument.  
Since $x^9 = e$ and $x^{11} = e$, the order of $x$ divides both $9$ and $11$.  Therefore, the order of $x$ is $1$ so $x = e$.
