elements of order $3$ in $A_n$

Question

Let $S_n$ be the permutation group on $n$. We know that $A_n \trianglelefteq S_n$.

How many elements $ a \in A_5$ have order three.

Is there any formula for finding number of elements in $ S_n $ or $A_n$ of order $k$? $\;(k \leq n)$

Answer 1

An element of $S_n$ is a product of commuting cycles. The order of a cycle is the length of the cycle. Thus the order of an element is the least common multiple of the lengths of the disjoint cycles. In $S_5$, therefore, the only elements of order 3 are the 3 cycles by length considerations.

A cycle is in $A_n$ if and only if it is of odd length. Thus all of the 3 cycles are in $A_n$. Given 3 positions, there are two 3 cycles that can fit in those positions. There are $\binom{5}{3}=10$ ways to choose 3 positions. Thus there are 20 elements of order 3.

Answer 2

The number of elements of order $3$ are only the $3$ cycles The $3$ cycles are a product of even number of transpositions and hence belong to $A_5$.Number of $3$ cycles =$\frac{1}{3}5P_3$

Answer 3

If $n \geq 6$, then $A_n$ has other elements of order $3$ than the $3$-cycles, too. That is, the products of disjoint $3$-cycles. The number of products of $k$ disjoint $3$-cycles in $A_n$ (where $n \geq 3k$) is $$ \frac{1}{3^kk!}\prod_{i = 0}^{3k-1}(n-i). $$ Then you take the sum of all those numbers.