Question: I am trying to solve a problem, not for homework, and it has me stomped!

> 
> For $n\geq 4$ and $\alpha\in S\_n$, $$\alpha=\dot{\alpha}\dot{\beta}$$ where $\dot{\alpha},\dot{\beta}$ are of order 2.
> 
> 
> 

I know that every permutation can be expressed as a product of transpositions.

In my attempt to solve the problem I have focused on cyclic permutations, as every permutation is a product of disjoint cyclic permutations.

Using the fact that, for $4<k\leq n$$$(a\_1\ a\_k)(a\_1\ a\_2\ \cdots\ a\_{k-1})=(a\_1\ a\_2\ \cdots\ a\_k),$$ I found that if $k$ is even then $$(a\_1\ a\_2\ \cdots\ a\_k)=[(a\_1\ a\_k)(a\_{k-1} a\_{k-2})\cdots (a\_{k-2i+1}\ a\_{k-2i})](a\_{k-2i-1}\ \cdots a\_{k-3}\ a\_{k-1})$$ where $k-2i+1=3$, and if $k$ is odd then $$(a\_1\ a\_2\ \cdots\ a\_k)=[(a\_1\ a\_k)(a\_{k-1} a\_{k-2})\cdots (a\_{k-2i+1}\ a\_{k-2i})](a\_{k-2i+1}\ \cdots a\_{k-3}\ a\_{k-1})$$ where $k-2i+1=2$.

I am wondering, is this the correct route to take in order to resolve this problem? Or, is there another way? Could someone point me in the right direction?

Answer: 
I think you are working in the right direction. Essentially, you build upon the following formulas (note that I compose permutations left-to-right)

$$
(12) \cdot (23) = (132)
$$

$$
(12)(34) \cdot (23) = (1342)
$$

$$
(12)(34) \cdot (23) (45) = (13542)
$$

$$
(12)(34) \dots \cdot (23) (45) \dots = (135 \dots 6 4 2).
$$
An explicit decomposition of the cycle $(12...n)$ is also given by the product of $r(12)r(34...n)$ and $r(123)r(4...n)$ where $r$ means order reversal, i.e. $r(123)$ maps $123$ to $321$. Clearly each factor is of order 2.