A 2-Sylow subgroup of $S_5$ is isomorphic to $D_4$ Here is what I've got so far: Let $H$ be a 2-Sylow subgroup of $S_5$. Since $|S_5|=120=2^3\cdot 3\cdot 5$, thus $|H|=8$. We also know that $D_4$=8. So it seems like I need to find a homomorphism $\phi$ the maps $H \rightarrow G$? Or am I on the wrong track?...
 A: There are five different groups of order $8$ up to isomorphism so just knowing the order is not enough. You need to actually find a $2$-Sylow subgroup, the others will be isomorphic to it by Sylow theorems. Let $r=(1234), s=(13)$. It is easy to check that $rs=sr^{-1}$, the order of $s$ is $2$ and the order of $r$ is $4$. Hence $\langle r,s\rangle$ is isomorphic to $D_4$. 
How did I guess it? Well, think about a square with vertices $1,2,3,4$, in that order clockwise. Now we can look for permutations that describe where does each vertex move after using an element of $D_4$ on that square. The permutation $(1234)$ describes where does each vertex move after a rotation by angle $\frac{2\pi}{4}$ clockwise, $(13)$ describes where does each vertex move after a reflection between two vertices. So using the definition of the dihedral group it makes sense to believe these permutations generate a group which is isomorphic to it-after all they describe the elements of $D_4$. 
A: The other way to think about this is to recall that all p-Sylow subgroups of symmetric groups are direct products of iterated wreath products of cyclic groups of order $p$ (the iterated wreath products correspond to the powers of $p$ present in the representation of $n$ in base $p$, with multiplicity).  In particular, this gives $2 Wr 2$, which is isomorphic to $2^2\rtimes 2$, which is isomorphic to $D_4$ as the 2-Sylow in both $S_4$ and $S_5$.
