Express the following as a product of two-cycles $((142)(3245))^{-1}$
My solution goes like this:
If $\beta=(142)(3245)$, then $\beta\in S_5$, as nothing is specified about the permutation. So,
$$\beta:1\longrightarrow 4, \\ 2\longrightarrow 2, \\ 3\longrightarrow 1, \\ 4\longrightarrow 5, \\ 5\longrightarrow 3.$$
Now, $$\beta^{-1}: 1\longrightarrow 3, \\ 2\longrightarrow 2, \\ 3\longrightarrow 5, \\ 4\longrightarrow 1, \\ 5\longrightarrow 4.$$
Thus, $$\begin{align} \beta^{-1}&=((142)(3245))^{-1}\\ &=(1354)(2)\\ &=(1354)\\ &=(13)(35)(54). \end{align}$$
We can omit a cycle of length $1$ while expressing a permutation a product of cycles(or transpositions) since cycles of length $1$, when multiplied with any permutation keeps the permutation (i.e permutation with which the single cycle of length $1$ is multiplied) unchanged or invariant.
Is the above solution correct (along with my reasonings)? If not, where is it going wrong?
By "socks and shoes": $$(gh)^{-1}=h^{-1}g^{-1},$$ we get $$\beta^{-1}=(3245)^{-1}(142)^{-1}=(4235)(241)=(4135)=(45)(43)(41).$$
You're solution is also correct.