Every permutation is a product of two permutations of order 2

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?

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

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

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

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

$$ (12)(34) \dots \cdot (23) (45) \dots = (135 \dots 6 4 2). $$

As an example of decomposing a cycle into two sets of transpositions:

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.