Product of two elements of order q Let $G$ be finite group. Let $x$ and $y$ be two elements of order a power of $q$, where $q$ is prime. Is the order of $xy$ equal to a power of $q$ (or of order 1)? thanks!
 A: Find a solved Rubik's cube.  The group of symmetries of the Rubik's cube is a very fun group which you can get your hands on.  Let R be the move which turns the right face clockwise by a quarter turn, and let U be the move which turns the top face clockwise by a quarter turn.  These both have order 4 which is a power of 2.  Compute the order of RU (that is start with a solved Rubik's cube and do RURURU... until you get back to a solved Rubik's cube).
A: As mentioned in the comments, the answer is no, and an example is given in another answer.
What fails is that we wish to compute $(xy)^n$ and compare this to $x^n$ and $y^n$. If $x$ and $y$ commute, this is fine, as we get $(xy)^n = x^ny^n$. But in general this does not hold.
We can, however, be somewhat more precise about how badly this fails, as this is given by the Hall-Petrescu formula (valid in any group, not just finite ones) which says that $$x^ny^n = (xy)^n\prod_{i=2}^nc_i^{\binom{n}{i}}$$ where each $c_i\in K_i(\left<x,y\right>)$ and where $K_i(H)$ (for some group $H$) is defined by $K_1(H) = H$, $K_{i+1}(H) = [H,K_i(H)]$ (where $[-,-]$ denotes the commutator).
So if we know something about the possible orders of elements in the various $K_i(\left<x,y\right>)$ then we can sometimes also say something about the order of $xy$ from knowing the orders of $x$ and $y$.
A: You may know that the symmetric group $S_n$ is generated by transpositions. If your conjecture were true, the order of all elements of $S_n$ would be a power of $2$.
