Let $G$ be a finite noncommutative group of order $N$, and let $x, y \in G$. Let $a$ and $b$ be the orders of $x$ and $y$, respectively. Can we say anything nontrivial about the order of $xy$ in terms of $a$ and $b$? If it helps, you may assume that $a$ and $b$ are coprime.

$\begingroup$ No, it is moreorless answered here: mathoverflow.net/questions/130697/… $\endgroup$ – Bugs Bunny Mar 31 at 6:57

$\begingroup$ Only in the comments, to be fair... $\endgroup$ – Bugs Bunny Mar 31 at 6:59

8$\begingroup$ To be more explicit: for all three positive integers $a,b,c$, there exist a finite group and elements $x,y$ such that $x,y,xy$ have order $a,b,c$, with the only trivial restriction: if one of $a,b,c$ is $1$, then the other two are equal. $\endgroup$ – YCor Mar 31 at 7:56

2$\begingroup$ Let me also mention that the case when $a,b,c$ are all even is trivial: one can then find three such elements in the direct product of three dihedral groups $D_a\times D_b\times D_c$, each with generators of order two $u,v$, namely $x=(uv,u,v)$, $y=(v,uv,u)$, $xy=(u,v,vu)$. $\endgroup$ – YCor Mar 31 at 10:35

$\begingroup$ Given $a,b,c\ge 2$, estimating/computing the smallest size $n$ such that there is such a pair in the symmetric group $S_n$, looks like an interesting problem. For $a,b,c$ even, the above gives an upper bound of $a+b+c$. $\endgroup$ – YCor Mar 31 at 10:42
The following theorem (which does not take the order $N$ of the group $G$ into account) shows that all possible combinations of $a$, $b$ and the order of $xy$ are possible. See Theorem 1.64 from Milne's course notes on group theory.
For any integers $a,b,c > 1$, there exists a finite group $G$ with elements $x$ and $y$ such that $x$ has order $a$, $y$ has order $b$, and $xy$ has order $c$.
I strongly suspect that the answer is "very little", unless you are looking for bounds on the order of $xy$ in terms of $N$ or something like that.
Example. Let's focus on the example $a = 2$ and $b = 3$. It is wellknown that \begin{align*} \mathbf Z/2 * \mathbf Z/3 &\stackrel\sim\to \operatorname{PSL}_2(\mathbf Z)\\ x &\mapsto \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix},\\ y &\mapsto \begin{pmatrix}0 & 1 \\ 1 & 1\end{pmatrix}, \end{align*} where $x$ and $y$ are the generators with $x^2 = 1$ and $y^3 = 1$. The product $xy$ maps to $$\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix},$$ which has infinite order. For any $n \in \mathbf N$, the surjection $$\operatorname{PSL}_2(\mathbf Z) \twoheadrightarrow \operatorname{PSL}_2(\mathbf Z/n)$$ gives a finite group of order roughly $n^3$ in which $xy$ has exact order $n$.
Example. Another example, still with $a = 2$ and $b = 3$, is the group $$G = \mathbf Z/n \wr S_3 = (\mathbf Z/n)^3 \rtimes S_3,$$ where multiplication is defined by $$(a_1,a_2,a_3,\sigma)(b_1,b_2,b_3,\tau) = (a_1+b_{\sigma(1)},a_2+b_{\sigma(2)},a_3+b_{\sigma(3)},\sigma\tau).$$ Then one easily checks that $x = (1,1,0,(12))$ has order $2$ and $y = (0,0,0,(123))$ has order $3$, and $xy = (1,1,0,(23))$ has order $2n$. This time we get any even number as the order of $xy$, inside a group of order $6n^3$.