# Order of product of group elements

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.

• No, it is more-or-less answered here: mathoverflow.net/questions/130697/… – Bugs Bunny Mar 31 at 6:57
• Only in the comments, to be fair... – Bugs Bunny Mar 31 at 6:59
• 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. – YCor Mar 31 at 7:56
• 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 $x=(uv,u,v)$, $y=(v,uv,u)$, $xy=(u,v,vu)$. – YCor Mar 31 at 10:35
• 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 $a+b+c$. – 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 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 $$6n^3$$.