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Main question: How do I calculate the number of $2$-sylow subgroups of $S_{2^n}$?

Let $n \in \mathbb{Z}_{\geq 2}$. I have a $2$-sylow subgroup $H \subset S_{2^n}$ (too long to spell out all the details of $H$) and I want to show it is its own normalizer. I think showing $|S_{2^n}: H| = |S_{2^n}:N(H)|$ would be a good idea, since I know the index of $H$.

By the sylow theorems we know that the index of $N(H)$ is exactly the number of $2$-sylow subgroups. However, I do not know exactly how to calculate this number.

On the wiki page, https://en.wikipedia.org/wiki/Symmetric_group#Sylow_subgroups, they say something about $W_p(n)$ and that it is the wreath product of $W_p(n-1)$ and $W_p(1)$. I think this could be useful if we can calculate $W_p(1)$. However I have never seen this notation and I cannot find anything online. Does anybody know a good source or just knows how to calculate the number of $2$-sylow sybgroups in $S_{2^n}$?

Rico
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  • It is not completely clear what you are asking. Are you asking for help in proving that $H = N(H)$? – Derek Holt Mar 30 '16 at 10:32
  • The question is the first sentence (calculating number of sylow subgroups). – Rico Mar 30 '16 at 10:50
  • Then just do what you suggest and prove that $H = N(H)$. That will prove that the number is $||S_{2^n}:H|$. I was unsure whether you needed help proving $H=N(H)$ or with calculating $|S_{2^n}:H|$. – Derek Holt Mar 30 '16 at 11:19
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    http://math.stackexchange.com/questions/1376586 is related. As you will see from the discussion, these proofs are moderately hard. – Derek Holt Mar 30 '16 at 11:22
  • I want to prove $H = N(H)$ by proving that the index of both groups are the same. But the answer in the related question proves it: I have $p=2$ so the number of $2$-sylow groups is precizely $|S_n|/|H|$. Probably quite hard but oh well. Thnx anyway – Rico Mar 30 '16 at 20:19

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