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I need to describe what are the $2$-Sylow subgroups of $S_{14}$. I don't know how to start thinking about this since is it too large. I know that all that I must do is to find one of them since all the others are conjugates, but it's not helping.

user2345215
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  • Can you find proper subgroups of $S_{14}$ that are still divisible by the same power of 2? That is, can you find subgroups of odd index? A minimal such one is a Sylow 2-subgroup, but you can make progress with any medium-ish example. A really good idea is to try and find odd index subgroups of the form $G\times H$. – Jack Schmidt Apr 03 '14 at 13:55
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    Maybe this could help: http://math.stackexchange.com/questions/156360/describe-the-structure-of-the-sylow-2-subgroups-of-the-symmetric-group-of-degr – Jérémy Blanc Apr 03 '14 at 15:57

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Here is an outline of finding Sylow $p$-subgroups of symmetric groups. A similar technique works for classical groups.

Proposition: If $H \leq G$ has index not divisible by $p$, then every Sylow $p$-subgroup of $H$ is a Sylow $p$-subgroup of $G$.

Here is an example odd index subgroup to help out: $S_2 \times S_{12} \leq S_{14}$ has index $\binom{14}{2} = 91$, odd. Hence a Sylow $2$-subgroup of $S_2 \times S_{12}$ is still a Sylow 2-subgroup of $S_{14}$.

Sylow $p$-subgroups of direct products are easy to understand. See for instance this question: Sylow p-subgroup of a direct product is product of Sylow p-subgroups of factors

Proposition: If $P$ is a Sylow $p$-subgroup of $G$, and $Q$ is a Sylow $p$-subgroup of $H$, then $P \times Q$ is a Sylow $p$-subgroup of $G \times H$.

Probably you can find the Sylow 2-subgroup of $S_2$. What about $S_{12}$? Well, now we look for an odd index subgroup, preferably of the form $G \times H$. Well $S_4 \times S_8 \leq S_{12}$ has index $\binom{12}{4}=495$, odd.

Hence you are just left finding Sylow $2$-subgroups of $S_4$ and $S_8$.

These are best understood using a different odd index subgroup, the wreath product.

Proposition: If $P$ is a Sylow $p$-subgroup of $G$, and $Q$ is a Sylow $p$-subgroup of $H$ with $H$ a permutation group acting on $X$, then $P \wr_X Q$ is a Sylow $p$-subgroup of $G \wr_X H$.

There is a group $S_n \wr S_m \leq S_{nm}$ of index $\frac{(nm)!}{m! (n!)^m}$. For example, $S_4 \wr S_2 \leq S_8$ has index $\frac{8!}{2! (4!)^2} = 35$, odd.

Hence you are just left finding Sylow $2$-subgroups of $S_4$. Probably you can do that without any help, but $S_2 \wr S_2 \leq S_4$ has index $\frac{4!}{2! (2!)^2} = 3$, odd. Hence you are just left finding Sylow $2$-subgroups of $S_2$, which is about as simple as this technique can make it.

Each of the propositions is proven using simple arithmetic.

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