$S_n$ the symmetric group. Give an example of two elements $f,g$ of $S_n$ of order $2$ s.t. their product $f \circ g$ is of order $n$.

Question

Question:

Let $n \geq 3$ be an odd number. Let $G$ be the group of all the bijections from $\mathbb{Z} / n \mathbb{Z}$ to $\mathbb{Z} / n \mathbb{Z}$. Give an exemple of two elements of $G$ of order $2$ such that their product is of order $n$. (Product means here composition of two functions.)

Answer:

First of all we can note that the group $G$ is equivalent to the symmetric group $S_n$ equipped with the law "$\circ$" (composition of functions) and $Card(S_n)=n!$.

After that I am stuck.
I have tried to find $f$ and $g$ such that they can be decomposed in a "product" of 2-cycle so this will ensure that the order of $f$ and $g$ is $2$. But then i do not succeed to get $f \circ g$ is one big n-cicle or in other word that $f \circ g$ is of order $n$.

Any help will be greatly appreciated and different solutions too.
Thank you.

Answer 1

For $n=3$, any two distinct elements of order $2$ will do.

For $n=4$, look at $g=(2,3)$, $f=(1,2)(3,4)$. Then $g\circ f=(1,3,4,2)$. For $n=6$, take $g=(2,3)(4,5)$, $f=(1,2)(3,4)(5,6)$, so $g\circ f=(1,3,5,6,4,2)$. You can probably see a pattern now.

For $n=5$, take $f=(1,2)(3,4)$, $g=(2,3)(4,5)$. Then $g\circ f=(1,3,5,4,2)$.

Can you take it from here?

Answer 2

Thank to Arturo Magidin (cf selected answer).

By induction.

$n=3$:
$g=(1;2), f=(2;3) \Rightarrow f \circ g = (1;3;2)$.
$f, g$ are order two has they are 2-cycle and $f \circ g$ is of order $3$ as it is an unique cycle of length $3$
More over we have the following pattern of the cycle that is respected: all the odd number from $1$ to $n$ increasing by jump of $+2$ and after all the even number from $n-1$ to $2$ decreasing by step of $-2$.
Schematically: $1 \rightarrow 3 \rightarrow 2 \rightarrow 1$

For $n$ odd:
$g_n=(1;2)(3;4)(5;6)...(n-2;n-1)$ thus $n$ is not swap.
$f_n=(2;3)(4;5)(6;7)...(n-1;n)$ thus $1$ is not swap.
$f_n \circ g_n $ is of order $n$ and his unique cycle looks like this: all the odd number from $1$ to $n$ increasing by jump of $+2$ and after all the even number from $n-1$ to $2$ decreasing by step of $-2$.
Schematically: $ 1 \rightarrow 3 \rightarrow 5 \rightarrow ... \rightarrow n \rightarrow n-1 \rightarrow n-3 \rightarrow ...\rightarrow 4 \rightarrow 2 \rightarrow 1 $

For $n+2$ ($n$ odd):
$g_{n+2}=(1;2)(3;4)(5;6)...(n-2;n-1)(n;n+1)=g_n(n;n+1)$ thus $n+2$ is not swap.
$f_{n+2}=(2;3)(4;5)(6;7)...(n-1;n)(n+1;n+2)=f_n(n+1;n+2)$ thus $1$ is not swap.
It is obvious that $f_{n+2}, g_{n+2}$ are of order two as this is the composition of 2-cycle.
$f_{n+2} \circ g_{n+2} = f_n(n+1;n+2)g_n(n;n+1)$ as $g_n$ and $(n;n+1)$ have by definition different support we can writte $f_n(n+1;n+2)(n;n+1)g_n$
Schematically: this give $ 1 \rightarrow 3 \rightarrow 5 \rightarrow ... \rightarrow n \rightarrow n+2 \rightarrow n+1 \rightarrow n-1 \rightarrow ...\rightarrow 4 \rightarrow 2 \rightarrow 1 $ adn this is an unique cycle of length $n+2$ so by property we get that the order is too $n$.

Q.E.D.

Answer 3

Let $\pi$ be an $n$-cycle, e.g. $\pi = (1;2;3;...;n)$ and $\tau$ an involution such that $\tau \cdot \pi \cdot \tau = \pi^{-1}$. E.g. $\tau = (1;n)(2;n-1)(3;n-2)...$. Now set $\omega = \tau \cdot \pi$. It is trivial to show that $\omega$ is another involution. Obviously, $\tau \cdot \omega = \pi$.