Orders of a symmetric group Consider the symmetric group $S_5$. I would like to find how many elements of $S_5$ are of order 5, and how many are of order 6. I would also like to determine what the maximum order of an element in this group would be.
Here is what I have so far: elements of order 5 are the 5-cycles. Elements of order 6 (since a 6-cycle is impossible for 5 elements) must have at least an even cycle and a cycle of length divisible by 3, and 2 + 3 = 5, so elements of order 5 are a combination of 2-cycles and 3-cycles.
How can I find the total count of such elements of order 5 and order 6, and the maximum order? I'm not entirely sure where to go from here.
 A: John's got you on the first part.  I'll answer your second question.
Facts:


*

*Every element of $S_n$ can be written as the product of disjoint cycles.

*The order of the product of disjoint cycles is the least common multiple of the orders of those cycles.  In other words, if $\pi=\pi_1\cdots\pi_n$, $o(\pi)=\mbox{lcm}\{o(\pi_1),\ldots,o(\pi_n)\}$.
So what's the maximum order of an element of $S_5$?

 When making a cycle in $S_5$, you only have $5$ numbers to work with (in cycle notation).  How many ways can you make disjoint cycles using only $5$ numbers?  What's the biggest $\text{lcm}$ you can find this way?

Now how about $S_n$?

 There's no easy closed form for this, but I'm sure you've guessed this 'formula': it's the maximum $\text{lcm}$ attainable from the integer partitions of $n$.  This is called Landau's function.

A: Since it's been proposed that a similar question be closed as a duplicate of this one, I'll derive the number of permutations corresponding to a given cycle type in more generality.
Consider a cycle structure with $j_k$ cycles of length $k$, with $\sum_kkj_k=n$, and imagine it written out as pairs of parentheses enclosing appropriate numbers of slots. Each of the $n!$ assignments of the numbers from $1$ to $n$ to the slots yields a permutation, and we want to know how many are distinct.
Within each cycle of length $k$, there are $k$ cyclic permutations of the entries that would yield the same cycle. Among $j_k$ cycles of length $k$, there are $j_k!$ permutations that merely rearrange cycles of the same length. Thus $\prod_kk^{j_k}j_k!$ assignments lead to the same permutation, so the number of distinct permutations is
$$\frac{n!}{\prod_kk^{j_k}j_k!}\;.$$
See also this Wikipedia section.
A: The possible elements of $S_5$ are given by the various partitions of $5$.  For example, $(i_1i_2)(i_3i_4i_5)$ is an element of $S_5$ corresponding to the partition $5=2+3$.  
To count the number of such elements, we have $5$ choices for $i_1$ after which we have $4$ choices for $i_2$.  We must then divide by $2$ because $(i_1i_2) = (i_2i_1)$.  Hence we have $(5)(4)/2 = 10$ possibilities for $(i_1i_2)$.  Once we have chosen $i_1$ and $i_2$, we then have $3$ choices for $i_3$, 2 choices for $i_4$ and one remaining choice for $i_5$ giving $6$ possibilities for $(i_3i_4i_5)$.  However we observe that, for example, $(345) = (534) = (453)$ (i.e. each three cycle yields two other equivalent 3-cycles).  Thus we have $6/3 = 2$ possibilities for $(i_3i_4i_5)$.
This gives $$\frac{(5)(4)}{2}\cdot 2 = 20$$ possibilities for elements of the form $(i_1i_2)(i_3i_4i_5)$.
Can you generalize this for other elements of the group?
