I'm having trouble with a question from Armstrong's Groups and Symmetry on symmetric groups. To be more precise, it's Exercise 6.7:
Show that, if $n \ge 4$, every element of $S_n$ can be written as a product of two permutations of order 2.
If it helps, the exercise suggested to begin with cyclic permutations, but I'm still stuck.
It was already asked here, but the answer pointed to a link which is no longer working.
Thanks for the attention.
First of all it should be noted that if we prove the assertion for cycles only then we're done. Indeed if the permutation $p$ has cycle decomposistion $p = c_1c_2c_3\ldots c_n$ and each cycle $c_i = r_is_i$ with $r_i, s_i$ of order $2$ then $r_i$ and $s_i$ commute with $r_j$ and $s_j$ with $i \neq j$ since they act on different points (like the $c_i$ do). So we can write $p = r_1s_1r_2s_2 \ldots r_ns_n$.In a first step we can move $s_1$ to the end giving $p = r_1r_2s_2\ldots r_ns_1s_n$. In a second step we can move $s_2$ to behind $s_1$ giving $p = r_1r_2r_3s_3\ldots r_ns_1s_2s_n$. In a finite number of analoguous steps we end up with $p = r_1r_2\ldots r_ns_1s_2\ldots s_n$. where $r_1r_2\ldots r_n$ and $s_1\ldots s_n$ have order $2$.
Now let $p = (i_1, i_2, \ldots,i_n)$ be a cycle of length $n$. I will call a rotation a move that can be compared to the movement of an airplane propeller, e.g. for a tuple $[1,2,3,4,5]$ the rotation will be $[5,4,3,2,1]$ this rotation is realized by the permutation $r = (1,5)(2,4)$, leaving the point $3$ fixed. Let $s$ be the rotation on the tuple with the first element discarded, in our example $[2,3,4,5]$ giving $[5,4,3,2]$ realized by the rotation $s = (2,5)(3,4)$. We can now realize visually that starting from a given point and applying a rotation $r$ and then a rotation $s$ then the points ends up with its successor. If necessary write the numbers on a strip of paper put a needle in the midpoint of the rotation, do $r$ replace the needle to the midpoint of the rotation $s$ and apply $s$ a convince yourself that you recovered the permutation as $p = rs$ with $r$ and $s$ of order $2$.