I'm working on the following problem:
Show that if $n$ is at least 4, every element of $S_n$ can be written as a product of two permutations, each of with has order 2.
Now I've got a proof, but my proof works for $n\geq 2$, so I'm sceptical about the validity.
Remark$\quad$ Note how for any product of cycles $E = (a,b)(c,d)…(e,f)$ the order of $E$ is 2 if $a\neq b\neq c\ldots$ ($E$
is a product of disjoint cycles with order 2.) Note also how $E_1E_2$
then has order $2$ for the same reason. It thus suffices to show that
$S_n$ can be written as a product of 2-cycles.
Proof$\quad$ Consider $x = (a,b) \in S_2$. In this case, the proof is trivial. So assuming any $x\in S_n$ can be written as a product of
2-cycles, we will show that any $y \in S_{n+1}$ can be written as a
product of 2-cycles.Let $y \in S_{n+1} = (y_1\;y_2\;…\;y_{n+1})$.
Then $y = x \cdot c$ for some $x\in S_n$ and $c$ a 2-cycle that
transposes $y_{n+1}$ to the $i$'th place in the permutation. Then
since $x$ can be written as a product of 2-cycles, so can $y$. This
completes the proof.
It'd be great if someone could point out any mistakes.
Best Answer
(1 2 3) = (1 3) x (2 3)
(1 2 3 4) = (1 4)(2 3) x (2 4)
(1 2 3 4 5) = (1 4)(2 3) x (2 4)(1 5)
(1 2 3 4 5 6) = (1 4)(2 3)(5 6) x (2 4)(1 5)
(1 2 3 4 5 6 7) = (1 4)(2 3)(5 7) x (2 4)(1 5)(6 7)
(1 2 3 4 5 6 7 8) = (1 4)(2 3)(5 8)(6 7) x (2 4)(1 5)(6 8)
(1 2 3 4 5 6 7 8 9) = (1 4)(2 3)(5 9)(6 8) x (2 4)(1 5)(6 9)(7 8)
(1 2 3 4 5 6 7 8 9 10) = (1 4)(2 3)(5 10)(6 9)(7 8) x (2 4)(1 5)(6 10)(7 9)
If $n \geq 6$ then the factorization of the cycle of length $n$ is derived from the factorization of the cycle of length $n-1$, increasing the number of transpositions in the first factor if $n$ is even, increasing the number of transpositions in the second factor if $n$ is odd.