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I'm asked to find the 2-Sylow subgroups of the symmetric group $S_3$ using a certain method that is to find the 2-sylow subgroup of $S_{2^k}$. This method is illustrated in I.N.Herstein's "TOPICS IN ALGEBRA" book. I understand the illustration in the book but don't know how to apply them to this problem since 6 is not a power of 2. I saw the solution to this problem but couldn't understand some points in it either. I'll give an outline of the solution that I saw.

Solution: divide $1,2,3,4,5,6$ in 3 clumps $\{1,2\},\{3 ,4\},\{5,6\}$

Then define $\sigma =(1 3)(2 4)$

Why?! In the book it's defined in the proof that $\sigma$ should be like $\sigma =(1 3 5)( 2 4 6)$ why do they change it into 2 cycles to this problem?! I thought that they consider $S_4$ and then later embed it into $S_6$ since 4 is a power of 2. But then again

The first 2- Sylow subgroup is defined as $$P_1 = \{( 1 2)\}$$ $$P_2 = \sigma^{-1}P_1\sigma =\{( 4 3)\}$$ $$P_3 = \sigma^{-2}P_1\sigma^2 =\{( 56)\}$$ How?! Do they choose $P_3$ or calculate it somehow?

It's given that the 2-sylow subgroup P in $S_8$ is given by $$P=\{( 1 2),( 56 ) , (1 3)(2 4)\}$$

Where did $P_3 $ go?!

Can anyone analyse and explain this. Or suggest any other way to prove this using the same method illustrated in the book as for $S_{p^k}$?!

Lakshmi Priya
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    Note that $S_n\times S_m$ embeds in $S_{m+n}$, and $\lvert S_{n+m}:S_n\times S_m\rvert$ is prime to $p$ when (conditions). – user10354138 Sep 09 '23 at 16:51
  • @user10354138 this problem is given before they say anything about direct products. Is there anyother way u could explain using sylow subgroup properties and permutation properties. – Lakshmi Priya Sep 09 '23 at 17:02
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    The $2$-Sylow subgroup of $S_6$ has order $8$; the fact that $6$ is "not a power of $2$" does not seem to be relevant. The rest of your questions are unintelligible without the book in front of me. "In the book it's defined in the proof that $\sigma=(135)(246)$". (Which would be an element of order $3$, and hence rather irrelevant for the purpose of $2$-Sylow subgroups...) "The" proof? Because there is only one? And once you define $\sigma$ once you are never allowed to define a new value of $\sigma$ in any context ever again? – Arturo Magidin Sep 09 '23 at 18:08
  • @ArturoMagidin your first claim is wrong. – Brauer Suzuki Sep 10 '23 at 05:59
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    @BrauerSuzuki Yeah, 16. Doesn't change the main point, though. – Arturo Magidin Sep 10 '23 at 17:42
  • @ArturoMagidin about $\sigma$, in the proof the general definition is given for $\sigma$ for any group $S_{p^k}$. But for this problem if i apply the general definition I'm getting a $\sigma$ of order 3. I don't understand how to choose $\sigma$ – Lakshmi Priya Sep 11 '23 at 02:37
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    In this old thread the Sylow 2-subgroup of $S_{22}$ is discussed. I hope that's illuminating enough. It goes like in the (+1) answer by Brauer Suzuki. – Jyrki Lahtonen Sep 11 '23 at 04:12
  • Here you see Sylow $p$-subgroup of $S_{p^2}$. – Jyrki Lahtonen Sep 11 '23 at 04:15
  • @JyrkiLahtonen this says a lot on my doubt... thanks!! – Lakshmi Priya Sep 11 '23 at 12:53

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Since no one wants to elaborate, I give it a try. Let $n=2^{a_1}+\ldots+2^{a_k}$ be the 2-adic expansion of $n$ with $0\le a_1<\ldots <a_k$. Let $P_1$ be a Sylow $2$-subgroup of $S_{2^{a_1}}$, let $P_2$ be a Sylow $2$-subgroup of $\mathrm{Sym}\{2^{a_1}+1,\ldots,2^{a_1}+2^{a_2}\}$ and so on. Check that $P=P_1\times\ldots\times P_k$ is a Sylow $2$-subgroup of $S_n$.

Brauer Suzuki
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  • "$\mathrm{Sym}{2^{a_1}+1,\ldots,2^{a_1}+2^{a_2}}$" does this mean the symmetric groups with n= $2^{a_1}+1,\ldots,2^{a_1}+2^{a_2} $ ? – Lakshmi Priya Sep 11 '23 at 02:17
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    The symmetric group on the set of integers ${2^{a_1}+1,\ldots,2^{a_1}+2^{a_2}}$. This group is isomorphic to $S_{2^{a_2}}$. – Brauer Suzuki Sep 11 '23 at 13:38