# 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.

It was already asked here, but the answer pointed to a link which is no longer working.

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