I was asked to decompose the permutation $$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5\\ 2 & 3 & 4 & 5 & 1 \\ \end{pmatrix} = (12345) \in S_5$$ into a product of two permutations, each with order 2.
I don't think that it is possible since the $\sigma$ is of order 5. But how can I show this mathematically?
On
The fact that $\sigma$ is of order $5$ doesn't imply that it can't be the product of two permutations of order $2$. That would only be the case if those permutations commute.
You can find two such permutations by trial and error. A permutation of order $2$ only has cycles of length $1$ (fixed points) and $2$ (transpositions). There are only two different types of transpositions with respect to the cycle $(12345)$, namely $(12)$ and $(13)$, so there aren't that many choices to try.
It is difficult to predict the order of the permutation $\sigma\rho$ just from the order of the permutations $\sigma$ and $\rho$. In this case, we may use $\sigma = (1 5)(2 4)$ and $\rho = (5 2)(4 3)$. Both clearly have order $2$ (being the product of two disjoint $2$-cycles), but we have $$ \sigma\rho = (1 5)(2 4)(5 2)(4 3) = (12345) $$ when read left-to-right. If you prefer reading these products right-to-left, then use $\rho\sigma$ instead.
(This was found by trial and error using WolframAlpha because I'm too lazy to find pen and paper, and I have internet available.)