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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!

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    I changed the title from "direct product" to "product" since that's what the question actually asks about. However, the original title suggests that you're confused about what "direct product" means, so I'm not completely sure that I made the right change here.2012-10-19
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    Hint: The answer is no. You can find easy examples in symmetric groups.2012-10-19

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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).

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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 $c_i\in K_i(\left)$ 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 $K_i(\left)$ then we can sometimes also say something about the order of $xy$ from knowing the orders of $x$ and $y$.

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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 $2$.