Find Sylow 7-Subgroup of S7 and how many are there?

Question

• Find a Sylow 7-subgroup of $S_7$. How many Sylow 7-subgroups of $S_7$ are there?

$|S_7|=7!= 7\cdot 6^2 \cdot 5 \cdot 2^2 $

$n_7 \equiv $ 1(mod 7)

• Find a Sylow 3-subgroup of $S_6$. How many Sylow 3-subgroups of $S_6$ are there?

$|S_6|=6!=3^2 \cdot 5 \cdot 2^4 $

So all Sylow 3-subgroups of $S_6$ are subgroups of order $3^2$.

Could someone provide a step by step answer of one of these questions, so I can attempt to do the other one? I've begun both but I'm not sure how to continue.

Answer 1

$\bullet$ Claim: The number of $p$-Sylow subgroups in the symmetric group $S_p$ is $(p - 2)!$.

Proof: Any $p$-Sylow subgroup is cyclic of order $p$ and has precisely $p-1$ generators. Moreover, if two $p$-Sylow subgroups share a generator, they are identical. So, the elements of order $p$ are partitioned according to which $p$-Sylow subgroup they belong to. We need to count the number of elements of order exactly $p$. This is precisely the number of distinct $p$-cycles, which is $\frac{p!}{p} = (p-1)!$. Grouping them into distinct $p$-Sylow subgroups (with $p-1$ in each stack), we see that the number of $p$-Sylow subgroups is $\frac{(p-1)!}{p-1} = (p-2)!$.

Corollary: There are $5!$ Sylow-$7$-subgroups in $S_7$, which are all conjugated and isomorphic to $C_7$.

Answer 2

So a Sylow 7-subgroup of $|S_7|$ is going to have order 7, by the prime decomposition you write down in your question.

Any group of prime order is cyclic, so all the Sylow subgroups are cyclic. Conversely, if you have a subgroup of order 7 then it's always going to be Sylow.

Therefore, you just need to identify the subgroups that are isomorphic to $C_7$. For this it suffices to identify all the elements of order 7, that is, the 7-cycles.

There are $7!/7 = 6!$ of these, and since each copy of $C_7$ has exactly one element not of order 7 (and no element of order 7 appears in two different copies of $C_7$) there are $6!/6 = 5!$ subgroups of order 7.

Answer 3

Here is a solution to the second question: The prime factorization of $6!$ takes the form $3^2m$, so a Sylow-3 group of $S_6$ has order $3^2 =9$. Use the following fact:

For any prime number $p$, groups of order $p^2$ are abelian and are either isomorphic to $\mathbb{Z}_{p^2}$ or to $\mathbb{Z}_p \times \mathbb{Z}_p$.

Since there are no elements of order 9 in $S_6$, any Sylow-3 subgroup must be generated by 2 disjoint 3-cycles. There are ${6}\choose{3}$$ = 20$ 3-cycles, so there are $20 / 2 = 10$ distinct disjoint pairs of these cycles.