I have the following problem
Find a finite group $G$ and $x, y \in G$, in which $|xy| > |x| |y|$.
Edit: I forgot the condition $\text{gcd(|x|, |y|})=1$. But it's too late to change the question now.
I am looking for any solution in general, but I would particularly be interested in a solution in $S_n$, since to me that seems to most natural place to look (since we know there has to be a solution in there).
My Attempt
So my final answer, after a lot of trial and error, came out to this:
$G =D_{10} \times S_5$, with $x = (s, (12))$ and $y = (sr, (345))$
Before this, I tried to find the answer in the familiar small non-abelian groups $D_8, Q_8, S_3$. When I saw this didn't work, I decided upon permutations. Again after a lot more trial and error, I came to this reasoning: suppose we work in $S_8$. The highest possible order of an element is $15$, and that would be a $3$ cycle with a $5$ cycle, using all the numbers $1-8$. So if we could get $|x|=2$ and $|y|=3$, that would work. But we have to use all $8$ numbers, so we can't just have a $2$ cycle and a $3$ cycle. Furthermore, from experience, I tried to make the permutations as "tangled" as possible, so I tried $(12)(34)(56)(78)$ times $(135)$. But this didn't work.
Best Answer
For every $n \geq 5$, the dihedral group of order $2n$ contains pairs of reflections $x$ and $y$ such that $xy$ is a rotation of order $n$. Since $D_{2n}$ can be embedded in $S_n$, this gives examples in $S_n$ for all $n \geq 5$.
Here's an explicit construction in $S_n$ of two elements of order $2$ such that their product had order $n$. Let $\sigma, \tau \in S_n$ be defined by $$\sigma(i) = n+1-i \quad \mbox{for all $i$, and}$$ $$\tau(i) = \left\{ \begin{array}{ll} 1 & \mbox{if $i = 1$}\\ n+2-i \quad &\mbox{if $i \neq 1$}\end{array} \right.$$ so $\sigma$ and $\tau$ both have order $2$.
Then $$\tau\circ\sigma(i) = \left\{ \begin{array}{ll} 1 & \mbox{if $i = n$}\\ i+1 \quad &\mbox{if $i \neq n$}\end{array} \right.$$ so $\tau\circ\sigma$ is an $n$-cycle and so has order $n$.