2
$\begingroup$

Question is to :

exhibit all sylow $3$ - subgroups of $S_4$

What i have done so far is :

Number of elements in symmetric group is $|S_4|=4.3.2=2^3.3$

number of elements of order $3$ in $S_4$ are $\frac{4.3.2}{3}=8$

As any sylow $3$ subgroup contain only two non identity elements and as there are $8$ elements of order $3$ we see that there should be $4$ sylow $3$ subgroups.

With not much difficulty I could see that sylow $3$ subgroups are

  • $\{(1)(2)(3)(4),(123),(132)\}$
  • $\{(1)(2)(3)(4),(134),(143)\}$
  • $\{(1)(2)(3)(4),(124),(142)\}$
  • $\{(1)(2)(3)(4),(234),(243)\}$

This was easy because I have only one factor of $3$ in $|S_4|$ and so the sylow $3$ subgroup is cylcic and it is not so difficult to take some element of order $3$ and generate the subgroup.

Now the question is what if the group $|S_n|$ has $3^2$ in its factorization may be for example $|S_6|=6.5.4.3.2.1=5.3^2.2^3$

I guess this does not give much to think about as any group of order $9$ is abelian and in this case it is not cyclic.

May be I would take Sylow $2$ subgroup case.. It is of order $8$ so there is a possibility of non abelian group too.

Now i have no idea how to construct a Sylow $2$ subgroup in this case as this is not so simple as that of $S_4$.

Even if i know one sylow subgroup what I would do is just conjugate it so that i would get all other groups. (This is what i would do which may not be a better idea).

For example,

Exhibit all sylow $2$ subgroups of $D_{12}$

We have $D_{12}=\{\langle a,b\rangle : a^6=b^2=1 ; ab=ba^{-1}\}$ and $|D_{12}|=12=2^2.3$

total number of elements of order $2$ in $D_{12}$ are $7$

So, for sure there would be more than one sylow $2$ subgroup as any sylow $2$ subgroup of $D_{12}$ contains only $4$ elements..

Now the question is how do i form a subgroup of $4$ elements in $D_{12}$ Arbitrary selection of two elements with order $2$ does not generate a group of order $4$.

Just by some sense, I picked up $a^3,b$ and considered $H=\{1,a^3,b,a^3b\}$

This selection was just by choice though it is somewhat natural..

Now how do i get all other sylow $2$ subgroups of $D_{12}$ :O

Only option I have is to consider conjugates of this subgroup....

I see that $gHg^{-1}=hHh^{-1}\Rightarrow h^{-1}gH(h^{-1}g)^{-1}=H$

But then Normalizer of $H$ in $G$ is $H$ (In this case which i am very sure of :D) thus we should get $h^{-1}g\in H$ i.e., I would not see $hHh^{-1}$ for which I have $h^{-1}g\in H$ and for which I have calculated $hHh^{-1}$

I mean I would not check for $hHh^{-1}$ where $h^{-1}g\in H$ and I have already calculated $gHg^{-1}$

  • $H=\{1,a^3,b,a^3b\}\Rightarrow aHa^{-1}=a\{1,a^3,b,a^3b\}a^{-1}=\{1,a^3,a^2b,a^5b\}$
  • $H=\{1,a^3,b,a^3b\}\Rightarrow a^2Ha^{-2}=a^2\{1,a^3,b,a^3b\}a^{-2}=\{1,a^3,a^4b,ab\}$

I can conclude at this stage that these $H,aHa^{-1},a^2Ha^{-2}$ are all three subgroups of $D_{12}$ just by counting number of elements of order $2$. But then I am doing in a hope that i could use this to work for a little bigger group.

  • As $a^3\in H$ I would not check for $a^3Ha^{-3}$
  • $a^4Ha^{-4}=aHa^{-1}$ as $a^{-1}a^4\in H$
  • $a^5Ha^{-5}=a^2Ha^{-2}$ as $a^{-2}a^5\in H$
  • As $b\in H$ I would not check for $bHb^{-1}$
  • $abH(ab)^{-1}=abHba^{-1}=aHa^{-1}$ so i do not check for $abH(ab)^{-1}$
  • $a^2bH(ab)^{-2}=a^2bHba^{-2}=a^2Ha^{-2}$ so i do not check for $a^2bH(ab)^{-2}$
  • $a^3bH(ab)^{-3}=a^3bHba^{-3}=a^3Ha^{-3}$ so i do not check for $a^3bH(ab)^{-3}$
  • $a^4bH(ab)^{-4}=a^4bHba^{-4}=a^4Ha^{-4}$ so i do not check for $a^4bH(ab)^{-4}$
  • $a^5bH(ab)^{-5}=a^5bHba^{-5}=a^5Ha^{-5}$ so i do not check for $a^5bH(ab)^{-5}$

By this i would say that i have found all sylow $2$ subgroups of $D_{12}$ namely :

  • $H_1=\{1,a^3,b,a^3b\}$
  • $H_2=\{1,a^3,a^2b,a^5b\}$
  • $H_3=\{1,a^3,a^4b,ab\}$

Questions i have concerned with this are :

  • How do you calculate sylow subgroups of some concrete groups (Any reference would be so helpful)
  • Once I find some sylow subgroup by chance how far should i go to find all other sylow subgroups (Counting would be one possibility for sure. I am hoping for some other way to proceed, if any)

Thank you :)

$\endgroup$
  • 2
    $\begingroup$ It would be hard to come up with a general strategy, but having found $P \in {\rm Syl}_p(G)$, I would try computing its normalizer $N=N_G(P)$. Then the number of Sylow $p$-subgroups is $|G:N|$, and you can find them all by conjugating $P$ by the elements of a transversal of $N$ in $G$. For Sylow subgroups of the symmetric group, one critical component is the normalizer in $S_{p^n}$ of a Sylow $p$-subgroup of $S_{p^n}$, so it would be important to come up with a general description of that. (But I don't know one off hand.) $\endgroup$ – Derek Holt Feb 11 '14 at 9:32
  • $\begingroup$ @DerekHolt : Oh yes... I remember number of sylow subgroups is $|G:N|$... I for got that so i do not claim that i understood it... I would try learning it again...That was so helpful and thank you :) $\endgroup$ – user87543 Feb 11 '14 at 9:35
  • $\begingroup$ You may enjoy this concrete description of Sylow subgroups of $S_n$. $\endgroup$ – Jyrki Lahtonen Feb 14 '14 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy