The Maximum possible order for an element $S_n$

You will have to write out the possible forms a given permutation (expressed as the product of disjoint cycles) can take, and then use the convenient fact that for disjoint cycles $\sigma_{1}, \dots, \sigma_{k} \in S_{n}$,

$$|\sigma_{1} \dots \sigma_{k}| = \textrm{lcm}(|\sigma_{1}|, \dots, |\sigma_{k}|).$$

For example, in $S_{5}$, you have (up to isomorphism) the following forms that a given permutation (written as the product of disjoint cycles) can take:

1. $(1 2 3 4 5)$
2. $(1 2 3)(4 5)$
3. $(1 2 3 4)$
4. $(1 2)(34)$
5. $(1 2 3)$
6. $(1 2)$

Then figure out which of the above forms will have the greatest order.

There is a sequence of values (of Landau's function, $g(n)$) that you can refer to for many values of $n$.

There is a known upper bound on the function:

$$g(n) < e^{n/e}.$$

A0000793: Landau's function g(n): largest order of permutation of n elements, Equivalently, largest lcm of partitions of n.

Consider the permutation $p = (1 2)(3 4 5)$. It is an element of $S_5$, but it has order 6. The "disjoint cycles" are $(1 2)$ and $(3 4 5)$, which have lengths of 2 and 3, respectively.

If you don't understand the "cycle notation" $(1 2)(3 4 5)$ leave a comment and I will explain further. The short version is that $(1 2)(3 4 5)$ is the permutation which sends $1\mapsto 2$, $2\mapsto 1$, $3\mapsto 4$, $4\mapsto 5$, and $5\mapsto 3$.