So I need to find the number of elements of order $15$ in $S_{13}$. As there are no $15$-cycles in $S_{13}$ so we are looking for some disjoint cycle representation, whose order is $15$. We need to count the number of cycles of type $(\text{a b c d e})(\text{f g h})$. For the first we have $\binom{13}5\times$something..etc. So I don't know how to effectively count those cycles. I have seen similar questions here, but still I am unable to count them properly that because I didn't understand the procedure completely. So can anyone show me how to count these permutations? I want to learn it. Thank you.
Find the number of elements of order $15$ in $S_{13}$?
3
$\begingroup$
group-theory
permutations
symmetric-groups
-
4Don't forget $(abcde)(fghij)(klm)$ and $(abcde)(fgh)(ijk)$. – 2017-02-13
-
0@Arthur Oh yes. Thank you. – 2017-02-13
-
0Sorry, I miscounted the last one; it was too long. I edited my comment. – 2017-02-13
1 Answers
0
Hint
For any $n\ge2$ and $2\le p\le n$, the number of $p-$cycles in $S_n$ is :
$$c(n,p)=\frac {p!}p \binom np=\frac{n!}{p\left(n-p\right)!}$$
So the number of $5-$cycles in $S_{13}$ is :
$$c(13,5)=30\,888$$ and the number of permutations of the form $(abcde)(fhg)$ is :
$$N_1=c(13,5)\,c(8,3)=3\,459\,456$$
This should help you to complete the calculation of the number of permutations of order 15 in $S_{13}$.