# Finding two group elements $x,y$ in a finite group such that $|xy|$ is greater than $|x| |y|$

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.

• What is the question? In my opinion, the easiest examples are two involutions in a large dihedral group. – verret Dec 31 '19 at 1:53
• @verret: "I am looking for any solution in general, but I would particularly be interested in a solution in $S_n$" – Ennar Dec 31 '19 at 1:54
• @Ennar but they already have their own answer. Anyway, in my opinion, the easiest examples are two "consecutive" involutions in a large dihedral group, which you can think of as "adjacent" reflections of a regular polygon. If you insist on working in $S_n$, then this translates to something like (say $n$ is even) $(1~n)(2~n-1)\cdots (n/2~n/2+1)$ and $(2~n)(3~n-1)\cdots (n/2~n/2+2)$ which both have order $2$, but their product is $(123\cdot n)$. – verret Dec 31 '19 at 2:00
• @verret, yes, and they are now asking for another solution - that's the question. – Ennar Dec 31 '19 at 2:01
• Ahh I forgot an important condition, which is $\text{gcd}(|x|, |y|) = 1$. Too late the change the question though :( – Blue Dec 31 '19 at 2:04

As you are probably aware, this could never happen in abelian group, since by commutativity we have $$|xy| = \operatorname{lcm}(|x|,|y|) \leq |x||y|.$$

Looking at $$S_n$$ works, and we can easily tell that $$n\geq 5$$, since $$|x|,|y|\geq 2$$ (otherwise at least one of $$x,y$$ is identity), so we need $$|xy|$$ at least $$5$$ (and that can't happen in $$S_4$$).

Let's look for that, we need $$x$$ and $$y$$ of order $$2$$, so $$x$$ and $$y$$ can consist of cycles of length at most $$2$$. We are trying to hit a permutation that consists of single cycle of length $$5$$, for example. Let's just randomly select $$x = (12)(34)(5)$$. Intuitively, $$y$$ should "mess up" the cycles of $$x$$ (or get the permutations "tangled", as you say), so let's pick $$y = (13)(45)(2)$$. We get that $$xy = (14532)$$, so $$|x||y| = 2\cdot 2 \leq 5 = |xy|$$.

I don't know if we there is something smarter that one could employ to avoid trial and error.

• Thanks! Ahh I'm sorry I forgot the condition $\text{gcd}(|x|, |y|) = 1$; but too late to change the question now, $+1$. Would you happen to know of an answer under this condition? – Blue Dec 31 '19 at 2:07
• @Blue, similar intuitive idea leads to $x = (123)(456)$, $y=(14)(57)$ that gives $xy = (1576423)$ in $S_7$, and I think you can't pull it off in $S_6$, considering possible length cycles, if I'm not mistaken. – Ennar Dec 31 '19 at 2:27
• Thanks! I will think about it – Blue Dec 31 '19 at 2:27

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 $$n$$.

• Thanks! Ahh I'm sorry I forgot the condition $\text{gcd}(|x|, |y|) = 1$; but too late to change the question now, $+1$. Would you happen to know of an answer under this condition? – Blue Dec 31 '19 at 2:06