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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.

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  • $\begingroup$ What is the question? In my opinion, the easiest examples are two involutions in a large dihedral group. $\endgroup$ – verret Dec 31 '19 at 1:53
  • $\begingroup$ @verret: "I am looking for any solution in general, but I would particularly be interested in a solution in $S_n$" $\endgroup$ – Ennar Dec 31 '19 at 1:54
  • $\begingroup$ @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 $(123\cdot n)$. $\endgroup$ – verret Dec 31 '19 at 2:00
  • $\begingroup$ @verret, yes, and they are now asking for another solution - that's the question. $\endgroup$ – Ennar Dec 31 '19 at 2:01
  • $\begingroup$ Ahh I forgot an important condition, which is $\text{gcd}(|x|, |y|) = 1$. Too late the change the question though :( $\endgroup$ – Blue Dec 31 '19 at 2:04
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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.

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  • $\begingroup$ 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? $\endgroup$ – Blue Dec 31 '19 at 2:07
  • $\begingroup$ @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. $\endgroup$ – Ennar Dec 31 '19 at 2:27
  • $\begingroup$ Thanks! I will think about it $\endgroup$ – Blue Dec 31 '19 at 2:27
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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 $n$.

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  • $\begingroup$ 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? $\endgroup$ – Blue Dec 31 '19 at 2:06

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