Sylow p subgroups of symmetric group of order (p+i)!

Question

Say I have a prime number $p$ and $0 \leq i \leq p-1$ and then I have the symmetric group $S_{p+i}$. I can use the Sylow theorem to show that there is a Sylow p subgroup of $S_{p+i}$, but I want to find how many. I know that $n_p | m$ when you write $ p* m = (p+i)!$ And I know that $n_p = [S_{p+i} : N_{S_{p+i}}(P)]$ where $P$ is the Sylow p subgroup, but I don't know how to find the normalizer. Can anyone give hints about this? preferably ones that don't use the fact $n_p \equiv 1 \mod p$

Answer 1

Since $i < p$, each Sylow $p$-subgroup is cyclic of prime order $p$, so it contains $p-1$ elements of order $p$, each of which has a cycle representation which is a single cycle of length $p$. Moreover, all such cycles are contained in some Sylow $p$-subgroup, and the Sylow $p$-subgroups intersect trivially.

There are $\displaystyle \frac{1}{p} \cdot \frac{(p+i)!}{i!}$ distinct $p$-cycles in $S_{p+i}$. Each Sylow $p$-subgroup contains $p-1$ of these $p$-cycles, each $p$-cycle is in exactly one Sylow $p$-subgroup. Therefore, there are $\displaystyle \frac{1}{p(p-1)}\cdot \frac{(p+i)!}{i!}$ Sylow $p$-subgroups.

We can confirm that this count is congruent to $1\mod p$ as follows. First, we can write $$\frac{(p+i)!}{i!} = (p+i)(p+i-1)\cdots(p+i-(p-1))$$ The right hand side is the product of $p$ consecutive integers which includes $p$ and $p-1$. Dividing by $p(p-1)$, we are left with $p-2$ integers. Working modulo $p$, each integer and its inverse both appear in this list (we have removed $p-1$ which is $-1$ mod $p$ and hence is its own inverse). Therefore the product is $1$. (This is essentially a variation of Wilson's theorem.)