Question is to find
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Sylow $p$ subgroups of $S_{2p}$ for odd prime $p$ and show that this is an abelian group of order $p^2$
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Sylow $p$ subgroups of $S_{p^2}$ for odd prime $p$ and show that this is a non abelian group of order $p^{p+1}$
what I have done so far is :
$|S_{2p}|=(2p)!=(2p)(2p-1)\cdots (2p-p)\cdots 4\cdot3\cdot2\cdot1$
with out giving proper justification I would say that $p$ occurs only twice in factorization of $(2p)!$..
$p$ occurs first time in $2p$ and second time in $(2p-p)=p$..
(I do not know how to make above justification look better)
So, sylow subgroup of $S_{2p}$ is of order $p^2$..
I was first thought of seeing for some example May be something like $S_6=S_{2.3}$
Order of sylow $3$ subgroup of $S_6$ is $9$
only possibilities for orders of elements are $3$..
If i take some two arbitrary elements of order $3$ there is a possibility that they do not end up nicely i mean some thing like
$$(123)(124)=(13)(42)$$ would happen..
I would take two elements of order $3$ and then end up with an element of order $2$ which would be a problem as a group of order $9$ can not have an element of order $2$
So, I would make sure that those elements do not create any mess and commutes nicely and the choice that comes to my mind is $\{(123),(456)\}$ and the group generated by them is of order $9$
$$\{\langle(123),(456)\rangle\}=\{(123)^i(456)^j : 0\leq i,j\leq 2\}$$
I would like to repeat the same in terms of $S_{2p}$ I would consider first half of $\{1,2,3,\cdots,2p-2,2p-1,2p\}$ as one cycle and second half as another cycle. I mean $\{(123\cdots,p-1,p);(p+1,p+2,\cdots,2p)\}$
These two cycles commutes nicely and do not create any mess giving rise to a subgroup of order $p^2$ namely :
$$\{\langle(123\cdots,p-1,p);(p+1,p+2,\cdots,2p)\rangle\}$$
$$=\{(123\cdots,p-1,p)^i(p+1,p+2,\cdots,2p)^j : 0\leq i,j\leq p-1\}$$
So, I guess this is good enough..
$|S_{p^2}|=(p^2)!=(p^2)(p^2-1)\cdots (p^2-p)\cdots (p^2-2p)\cdots (p^2-(p-1)p) \cdots 4\cdot3\cdot2\cdot 1$
i.e., $|S_{p^2}|=p^2\cdot p(p-1)\cdot p(p-2)\cdots p(1)m$ where $p\nmid m$
i.e., we have $p^2\cdot p^{p-1}=p{p+1}$ number of $p$ in factorization of $(p^2)!$
So, now the order is known and we need to find the subgroup…
I would repeat the same idea as of $S_{2p}$
For the same reason as that of $S_{2p}$ I would take for $S_{9}=S_{3^2}$the set $\{(123),(456),(789)\}$.
But then $\{\langle (123),(456),(789)\rangle\}=\{(123)^i(456)^j(789)^k : 0\leq i,j,k\leq 2\}$ is giving me a subgroup of order $27$ where as i need a group of order $81$ (which is the order of sylow $3$ subgroup of $S_{9}$ i.e.$3^{3+1}$)
so, It sounds some thing like I should take an element of order $3$ which behaves nicely and does not create much problem and give a subgroup of order $81$
I am sure that any $3$ sylow subgroup of $S_9$ would have a copy isomorphic to $\{\langle (123),(456),(789)\rangle\}$ so only problem is i am not sure how to extend this to a subgroup of order $81$
More generally for $S_{p^2}$ I would consider
$$\{\langle(123\cdots p)(p+1\cdots 2p)\cdots (p^2-p+1\cdots p^2)\rangle\}$$
This would give a subgroup of order $p^p$ but then i need a subgroup of order $p^{p+1}$.
I would be thankful if some one can help me to see what would be the obvious element that one should multiply this with :O
Please help me to clear this.
Thank you.
Best Answer
To find Sylow $p$-subgroups in a group in which you can generally find $p$-elements and normalizers the following algorithm is sometimes used:
Assume $P_i$ is a $p$-subgroup of $G$ (we can take $P_0 = \{1_G\}$ to be the trivial subgroup). If $P_i$ is a Sylow $p$-subgroup, then yay, return $P=P_i$. Otherwise we know by theory* that $P_i$ is not a Sylow $p$-subgroup of $N_G(P_i)$ either. Let $x_i$ be any $p$-element of $N_G(P_i)$ not contained in $P_i$, and set $P_{i+1} = \langle x_i, P_i \rangle$. This is still a $p$-subgroup, since* $x_i$ normalizes $P_i$. Hence we can repeat.
Theory*: If $P_i$ is a not a Sylow $p$-subgroup, then it is properly contained in an as-yet-unknown Sylow $p$-subgroup $P$. Since $P$ is nilpotent, it satisfies the normalizer condition and so $P_1$ is a proper subgroup of the $p$-subgroup $N_P(P_1) \leq N_G(P_1)$. Hence $P_1$ is not a Sylow $p$-subgroup of $N_G(P)$.
Since*: If $H,K \leq G$ and $H$ normalizes $K$, then $HK=KH$ is a subgroup and $|HK| \cdot |H \cap K| = |H| |K|$. In our case these numbers are all powers of $p$.
Application to the symmetric group: You already have $$P_p = \left\langle (1,2,3,\dots,p), (p+1,p+2,p+3,\dots,2p),\dots,(p^2-p+1,p^2-p+2,\dots,p^2-1,p^2) \right\rangle$$ so you just need to find some element that normalizes this group. Hence you need some order $p$ automorphism of the group, say one that cycles the basis elements $$x_i=(p(i-1)+1,p(i-2),\dots,pi)$$ amongst themselves, cyclically, in a cycle of length $p$. Such an element is exactly $$x_p = (1,p+1,2p+1,\dots,p^2-p+1)(2,p+2,2p+2,\dots,p^2-p+2)\cdots(p,2p,3p,\dots,p^2)$$
Generally speaking normalizers can be hard to find, so this algorithm is only used when the more advanced algorithms are not available (which unfortunately is still pretty often AFAIK). However, it just works fine here.