Why are there elements of order $6$ in the permutation group $S_5$?

Question

I understand that an element in $S_5$ can have an order $6$ if it is product of two disjoint cycles of one of length $2$ and another of length $3$, but I do not understand why these elements have an order of $6$ since there are not any cycles in $S_5$ of length $6$. Please explain how this is possible.

Answer 1

Use the following theorem.

Theorem (Order of a Permutation) : The order of a permutation of a finite set written in disjoint cycle form is the least common multiple of the lengths of the cycles.

Answer 2

It is sufficient to provide one example. Consider

$$\sigma=(123)(45).$$

Then

$$\begin{align} \sigma^2&=(132),\\ \sigma^3&=(45),\\ \sigma^4&=(123),\\ \sigma^5&=(132)(45),\\ \sigma^6&={\rm id}_{S_5} \end{align}$$