I am trying to find the number of Sylow 2-sbgroups of the symmetric group S5. As $ \lvert S_5 \rvert =120=2^3 \cdot 3 \cdot 5$. It has 2-SSG, 3-SSG, 5-SSG. But how to calculate there numbers? What is the general formula to calculate it.
[Math] the number of Sylow 2-subgroups of the symmetric group $S_5$
abstract-algebrafinite-groupsgroup-theory
Related Solutions
I think the $S_5$ work is correct (sorry, haven't looked at the $A_5$ work). I would do it a somewhat different way:
Since 5, but not 25, divides 120 (the size of $S_5$), the Sylow-5 subgroups of $S_5$ must be cyclic of order 5. There are 24 5-cycles in $S_5$, 4 of them in each of these subgroups, so, 6 Sylow-5 subgroups.
Similarly, the Sylow-3 subgroups must be cyclic of order 3. There are 20 3-cycles in $S_5$, 2 to a subgroup, so 10 Sylow-3 subgroups.
Since 8, but not 16, divides 120, the Sylow-2 subgroups must have order 8. Now, $S_4$ contains three copies of the dihedral group of order 8, and $S_5$ contains 5 copies of $S_4$, so I get 15 Sylow-2 subgroups.
Something like this ought to work for $A_5$.
EDIT: I think OP wants me to elaborate on the dihedral-group part of the argument.
Take a square, label its vertices, cyclically, with 1, 2, 3, 4. Then the element $(1234)$ of $S_4$ has a natural interpretation as the rotation, one-fourth of the way around, of the square, and the element $(13)$ is the flip in the diagonal through 2 and 4, so these two elements of $S_4$ generate a subgroup isomorphic to the dihedral group of order 8.
The same is true for the elements $(1342)$ and $(14)$, and also for the elements $(1423)$ and $(12)$, and those are the three copies of the dihedral group in $S_4$.
Now if you pick any one of the numbers 1, 2, 3, 4, 5, and consider all the elements of $S_5$ that fix that number, you get a subgroup of $S_5$ isomorphic to $S_4$. Those are your five copies of $S_4$ in $S_5$.
There's not a simple formula known for this, and certain aspects of the question are the subject of ongoing research. For example, this paper summarizes some research that has been done on the maximum possible order for an element of $S_n$.
In general, the way to find the number of elements of order $k$ in $S_n$ is:
Determine all possible cycle types for an element of order $k$, and then
Determine the number of elements having each of these cycle types.
For step (1), you're just looking for all possible ways to partition $n$ into cycles so that the least common multiple of the cycle lengths is $k$. For example, if permutation has order six, then all the cycles must have length $1$, $2$, $3$, or $6$, with either at least one $6$-cycle or one cycle each of lengths $2$ and $3$. So if we want to count the number of permutations of order six in $S_8$, the possibilities are
- One $6$-cycle, one $2$-cycle,
- One $6$-cycle, two $1$-cycles,
- Two $3$-cycles, one $2$-cycle,
- One $3$-cycle, two $2$-cycles, and one $1$-cycle, or
- One $3$-cycle, one $2$-cycle, and three $1$-cycles.
Step (2) is easy once you figure out step (1). In particular, the number of permutations in $S_n$ with a given cycle structure is $$ \frac{n!}{\prod_{d=1}^n (c_d)!\,d^{c_d}} $$ where $c_d$ denotes the number of cycles of length $d$. For example, the number of elements of $S_{20}$ having four $1$-cycles, five $2$-cycles, and two $3$-cycles is $$ \frac{20!}{(4!\cdot 1^4)(5!\cdot 2^5)(2!\cdot 3^2)} \;=\; 1\text{,}466\text{,}593\text{,}128\text{,}000. $$ The following table shows the number of elements of each order in $S_2$ through $S_8$. $$ \begin{array}{crrrrrrrrrrr} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 10 & 12 & 15 \\ \hline S_2 & 1 & 1 & - & - & - & - & - & - & - & - & -\\ S_3 & 1 & 3 & 2 & - & - & - & - & - & - & - & - \\ S_4 & 1 & 9 & 8 & 6 & - & - & - & - & - & - & - \\ S_5 & 1 & 25 & 20 & 30 & 24 & 20 & - & - & - & - & - \\ S_6 & 1 & 75 & 80 & 180 & 144 & 240 & - & - & - & - & - \\ S_7 & 1 & 231 & 350 & 840 & 504 & 1470 & 720 & - & 504 & 420 & - \\ S_8 & 1 & 763 & 1232 & 5460 & 1344 & 10640 & 5760 & 5040 & 4032 & 3360 & 2688 \end{array} $$ This table is entry A057731 at OEIS.
Best Answer
There are $15$ Sylow-$2$-subgroups of $S_5$. For a proof see here. By Sylow's theorem we have $n_2(S_5)\in \{ 1,3,5,15\}$, and it is easy to show that $n_2(S_5)\ge 6$, so the result follows.