I'm trying to find how many elements in $S_9$ have order $7$.
Using the fact the order of $g$ is the lowest common multiple of the length of its disjoint cycles, the only combination of cycles possible is: one of length $7$, and two of length $1$.
What I'm having trouble with is finding how many $g$ ϵ $S_9$ are in this form.
My logic would be you would have $9$ options to pick for the "$1$ cycle" and then $8$ options for the next "$1$ cycle" and for the "$7$ cycle" there would be $6!$ options as each one cannot map to the starting elements.
So the number of elements in $S_9$ that have order $7$ would be $9*8*6! =51840$ yet I'm not sure this is correct because I have seen someone post a general formula for a similar problem (link below) that does not yield the same answer (It actually gives half of my answer, $25920$), can someone please point out what I'm missing?
Thanks.
Finding the number of elements of particular order in the symmetric group
Best Answer
The order of the two $1$-cycles doesn‘t matter, so you need to divide by $2!=2$.