Can someone explain to me why the symmetric group EDIT: $S_6$ has 10 Sylow 3-subgroups? I gather the Sylow 3-subgroups of $S_6$ are isomorphic to $\mathbb Z_3 \times\mathbb Z_3$ and that it must be of order 9, by prime decomposition. I thought I could use the binomial formula $ {6 \choose 3}$ to work out the number of sylow 3-subgroups, but of course this gives me 20.

I realise this is probably a simple question but any help would be appreciated.

  • $\begingroup$ Did you mean in $S_6$? Because that is correct for $S_6$, but incorrect for $S_7$. $\endgroup$ – Ken Duna Dec 2 '16 at 16:10
  • $\begingroup$ Yes, sorry. My mistake. $\endgroup$ – user377174 Dec 2 '16 at 16:13
  • $\begingroup$ That's the way to go about it. I'm on my way to class right now, but once I get my students started on their quiz, I can explain. $\endgroup$ – Ken Duna Dec 2 '16 at 16:29
  • $\begingroup$ Hint: $\langle (123), (456) \rangle$ is a Sylow $3$ subgroup. Note that there are two distinguished subsets of $\{1, 2, 3, 4, 5, 6\}$ of size $3$ in there. $\endgroup$ – Dustan Levenstein Dec 2 '16 at 16:38
  • $\begingroup$ Is that why you would divide 20 by 2? $\endgroup$ – user377174 Dec 2 '16 at 17:08

The $3$-Sylow subgroups are generated by pairs of disjoint $3$-cycles (this is where the $\mathbb{Z}_3 \times \mathbb{Z}_3$ comes from). How many ways can you choose a pair of disjoint $3$-cycles? You choose a $3$-cycle, and then its partner is already determined. So far there are $\binom{6}{3}$ ways to do this.

However, we have double counted. If instead, we had chosen our cycle's partner, then we would get the same two generators. This is why you divide by $2$.

  • $\begingroup$ Thank you for your answer :) Can I ask under what conditions can I use the binomial formula to find the number of p-sylow groups? For example, I can't use it to find the 2-sylow subgroups of $S_5$ can I? $\endgroup$ – user377174 Dec 2 '16 at 21:06
  • 2
    $\begingroup$ There is no "silver bullet" unfortunately. In the case of $2$-Sylows in $S_5$, the Sylow Theorems tell us that the number of $2$-Sylow subgroups must be odd and divide $15$. So there are $1,3,5,$ or $15$. One option to prove that $15$ is correct is to list at least $6$ of them. $\langle(1 \ 2 \ 3 \ 5), (2 \ 5)\rangle$ is one of them. You can get others by conjugating by elements of $S_5$. $\endgroup$ – Ken Duna Dec 2 '16 at 22:21

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