$2$-Sylow subgroups of $S_{14}$ 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.
 A: 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.
