Cycle decompositions of $S_{4}$?

Question

Can someone please explain how to form the cycle decompositions of $S_{4}$, more specifically those that have order 2?

I understand where $(1 2),(1 3),(1 4),(2 3),(2 4),(3 4)$, come from.

But I don't understand how to get the others such as $(1 2)(3 4)$.

Answer 1

A 2-cycle has period 2, so $(ab)(ab) = (1)$ (we interchange $a$ and $b$, and then again, so we are left with the identity).

If $x = (ab)(cd)$ where the cycles are disjoint (!), then we can interchange the multiplication between disjoint cycles, so $(ab)(cd)(ab)(cd) = (ab)(ab)(cd)(cd) = 1$ as well. So in any $S_n$, any product of disjoint 2-cycles has order 2 as well.

Hence $(12)(34)$, $(13)(24)$ and $(14)(23)$ are other elements of order 2.

Any permutation in $S_n$ can be written uniquely as some product of disjoint cycles (of different lengths), and if it contains some $3$-cycle it has at least order $3$, and if it has a $4$-cycle then it has order at least 4, etc. So the only elements of $S_4$ of order 2 are the $2$-cycles and the products of two disjoint $2$-cycles.