TITLE: Order of product of group elements
QUESTION [5 upvotes]: Let $G$ be a finite non-commutative 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 non-trivial about the order of $xy$ in terms of $a$ and $b$? If it helps, you may assume that $a$ and $b$ are coprime.

REPLY [20 votes]: 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 $c$.

REPLY [5 votes]: 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 well-known 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 $6n^3$.