Determine the structure of a Sylow $2$ subgroup of Sym($4$). Determine the structure of a Sylow $2$ subgroup of Sym($4$).
I have got so far as saying there are $3$ Sylow $2$ subgroups, but i do not know how to determine the structure.
Thanks
 A: You have gotten as far as deducing that there are $3$ Sylow $2$-subgroups of $S_4$.  The largest divisor of $24$ which is a power of $2$ is $8$, so each of the three Sylow $2$-subgroups have order $8$.
One easy way of determining the structure of these subgroups is by first recognizing that $S_4 \cong Cube$, the group of symmetries of a cube.  Next, define $S$ to be the set of pairs of opposite faces of a cube, and let $Cube$ act on $S$.  Note that $|S| = 3$ since the cube has $3$ pairs of opposite faces.  Now we can choose some $s \in S$ and apply the Orbit-Stabilizer theorem:
$$|Cube| = |G_s||\mathcal{O}_s|$$
Where $G_s$ is the stabilizer set of $s$ and $\mathcal{O}_s$ is the orbit of $s$.  Well, under $Cube$, any pair of opposing faces can be sent to any other pair, so $|\mathcal{O}_s| = |S| = 3$.  We also know $|Cube| = 24$, so $G_s = 8$ for all $s \in S$.  The stabilizer set $G_s$ is always a subgroup, and since $|S| = 3$, we therefore have isolated $3$ subgroups of order $8$.  At this point, we want to try to understand the structure of these stabilizer subgroups.  
To do this, we want to look at some $s \in S$, that is, a particular pair of opposite faces on a cube.  What elements of $Cube$ preserve this pair of faces?  I.e., what $g \in Cube$ are such that this pair is not getting sent to any other pair?  Well, the identity is one.  We also have $3$ rotations about the axis through the center of the faces.  Finally, we have the $180^\circ$ rotation that takes one of the faces in the pair to the other face.  If we take these together with compositions, we have $8$ elements of $G_s$.  Now what is this subgroup $G_s$ actually isomorphic to?
The choices are $\mathbb{Z}_8$, $\mathbb{Z}_4 \times \mathbb{Z}_2$, $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$, $D_4$, and the quaternion group.  Most of these should be fairly easy to eliminate right off the bat.
Certainly, it cannot be $\mathbb{Z}_8$ since there is no element of order $8$ in this subgruop.  Similarly, it cannot be $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ since there is an element of order $4$ in $G_s$. It cannot be the quaternion group since the quaternions have $3$ elements of order $4$, and $G_s$ does not. 
So our only choices left are $D_4$ and $\mathbb{Z}_4 \times \mathbb{Z}_2$.  To finish up, I recommend that you get a cube and play with the actions that preserve a pair of opposite faces.  If the subgroup is commutative, it will be $\mathbb{Z}_4 \times \mathbb{Z}_2$.  If it is not, then it will be $D_4$.
A: The Sylow $2$-subgroups of $S_4$ have size $8$ and the number of Sylow $2$-subgroups is odd and divides $3$. Counting shows that $S_4$ has $16$ elements of order dividing $8$, and since every $2$-subgroup is contained in a Sylow $2$-subgroup, there cannot be only one Sylow $2$-subgroup. More counting tells you that $S_4$ contains six $2$-cycles, three $2 \times 2$-cycles, and six $4$-cycles. Since the three Sylow $2$-subgroups of $S_4$ are conjugate, the different cycle types must be distributed “evenly” among the three Sylow $2$-subgroups. Since we have three Sylow $2$-subgroups and only $16$ elements to fill them with, any two Sylow $2$-subgroups must intersect. Moreover, we know exactly how they intersect: recall that the Klein $4$-group $V_4$ is normal in $S_4$, since it is a union of conjugacy classes. Therefore, each Sylow $2$-subgroup contains $V_4$ (since in particular some Sylow $2$-subgroup does and all Sylow $2$-subgroups are conjugate). We have determined that each Sylow $2$-subgroup contains two (disjoint) $2$-cycles, all three $2 \times 2$-cycles, and two $4$-cycles (each squaring to one of the $2 \times 2$-cycles and obviously inverses of each other). Therefore, the $3$ Sylow $2$-subgroups of $S_4$ are
$$P_1=\{(1),(12),(34),(12)(34),(13)(24),(14)(23),(1324),(1423)\}$$
$$P_2=\{(1),(13),(24),(12)(34),(13)(24),(14)(23),(1234),(1432)\}$$
$$P_3=\{(1),(14),(23),(12)(34),(13)(24),(14)(23),(1342),(1243)\}.$$
All of them are isomorphic to the dihedral group $D_4$.
