Writing every element of $S_n$ as a product of permutations of order 2

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.

Thanks for the attention.

• Hmm. I'm sure there's an algebraic proof of this but I'd like to see just a clever swapping one. I can do it for even-length cycles of length at least 4: e.g. (abcdef) = (ab)(cd)(ef) * (bc)(de), and I think for a simple transposition, you need something like (12) = (12)(34) * (34) hence $n\ge 4$. And of course for cycles of length 3: e.g. (abc) = (ab)(bc). Then for cycles of length 5, that was a bit trickier but (abcde) = (ab)(ce) * (be)(cd) seems to work. So with some cleaning, and remembering to put a single transposition into only one of the two elements used, I think this works. – Elizabeth S. Q. Goodman Apr 3 '17 at 19:55
• Please fix the title and write "as a product of permutations", otherwise it does not make sense: for each $n>2$, there are elements of order different from 2 in $S_n$. – Alexander Konovalov Apr 3 '17 at 20:08
• In the dihedral group of order $2n$ a rotation is a cycle of length $n$, and it can be written as the product of two reflections. – Derek Holt Apr 3 '17 at 20:18
• Edited the title and linked the previous question with the broken link for reference. Thanks for the tip! – AspiringMathematician Apr 4 '17 at 0:29

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 propellor, e.g. for a tuple $[1,2,3,4,5]$ the rotation will be $[5,4,3,5,1]$ this rotation is realized by the permutation $r = (1,5)(2,4)$, leving the point $3$ fixed. Let $s$ be the rotation on the tuple with the first element discarded, in our examploe $[2,3,4,5]$ 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 $2$.
• Before accepting the answer, I'd like to ask for a clarification on how to generalize the latter paragraph for larger $n$. Am I right if I say that every permutation can be found by using rotations (and excluding the right elements from them), and then I can commute those rotations,so I use the first paragraph? – AspiringMathematician Apr 4 '17 at 0:46
• If you give me an example I will gladly work it out for you. The rotations work for any cycle of length $n$.and combinations of them. – Marc Bogaerts Apr 4 '17 at 10:57