Permutation Group-$S_{10}$ How many elements of order $30$ are there in the symmetric group $S_{10}$?
I worked out and got $10500$ - using Computations and adjusting each individual cycle's position ( $2-3-5$ , $3-2-5$ etc). but then I realized that disjoint cycles commute. So got a little bit confused, Help? :)
 A: The only way to get an element of order $30$ in $S_{10}$ is as a disjoint product of a $2$-cycle, a $3$-cycle, and a $5$-cycle. So first we select the indices going into each category; the number of such choices is $$\frac{10!}{2! \cdot 3! \cdot 5!}.$$ Then we need to choose exactly what order to permute the indices for each cycle in. For the $2$-cycle, there is only one option. For the $3$-cycle, there are two choices. For the $5$-cycle, there are $4!$ choices (can you figure out why?). Multiply that out, and you should get a total of $120960$ elements of order $30$, if I did my math right.
A: We want to determine the number of permutations in $S_{10}$ of order 30.  Since the order of a permutation is the least common multiple of the lengths of its cycles, and since 30 is $2 \times 3 \times 5$ or $5 \times 6$ (but a 5-cycle and 6-cycle require 11 points, whereas we are in $S_{10}$), we want to determine the number of permutations in $S_{10}$ consisting of a 2-cycle, a 3-cycle, and a 5-cycle.  Here are two ways to do this:
We need to fill in the blanks in $(-,-)(-,-,-)(-,-,-,-,-)$.  The points in the first 2-cycle can be chosen in ${10 \choose 2}$ ways.  The next 3-cycle can be chosen in ${8 \choose 3} \times (3-1)!$ ways since in general there are $(n-1)!$ different circular permutations of length $n$.  Finally, the 5-cycle can be chosen in $(5-1)!$ ways.  Multiplying these out, we get ${10 \choose 2} \times 2 \times {8 \choose 3} \times 2! \times 4! = 120960$.
Another way is as follows.  The 10 blanks can be filled in $10!$ ways.  Consider the  blanks in the 3-cycle: the three choices $(1,2,3)$ and two of its circular shifts $(2,3,1)$ and $(3,1,2)$ are actually the same permutation (i.e. the same bijective function). So we need to divide by $3$ for this reason.  Overall, we need to divide $10!$ by the length of each of the cycles, and we get $\frac{10!}{2 \times 3 \times 5}=120960$.
