In $S_{4}$, find a Sylow 2-subgroup and a Sylow 3-subgroup. 
In $S_{4}$, find a Sylow 2-subgroup and a Sylow 3-subgroup.

With everyone's comments and inputs, I have outlined the following answer:
We have $|S_{4}|= 24 = 2^{3}3$. If $P$ is a Sylow $2$-subgroup then $|P| = 2^3$, and if $K$ is a Sylow $3$-subgroup then $|K|=3$. So we need to find subgroups of $S_4$ of order $8$ and order $3$. For the subgroup of order $3$, since it is of prime order, it is cyclic, thus we need to find an element of $S_4$ of order $3$.  So any $3$-cycle of $S_4$ will suffice.  As a concrete example, we will call $K = \langle(1\ 2\ 3)\rangle$.
Now we must find $P$. Since $|P| = 8$, every element in $P$ must have order $1$, $2$ or $4$.  We can choose a dihedral subgroup $D_4$ to be $P$. For example
$$P= \lbrace e, (1234), (13)(24), (1432),(24) ,(14)(23), (13), (12)(34)\rbrace.$$
 A: A few tips for finding a Sylow $2$-subgroup without too much manual labor:


*

*Every element of $S_4$ which is not a $3$-cycle has order $1$, $2$, or $4$, hence is contained in some Sylow $2$-subgroup. 

*By the Sylow theorems, $n_2$ is either $1$ or $3$. We can exclude $1$ as a possibility because we have more than $8$ elements of order $2$ or $4$.

*The Sylow $2$-subgroups are conjugate to each other, hence isomorphic to each other, so they each have the same number of elements of each cycle type.

*There are six transpositions ($2$-cycles) in $S_4$. Moreover, we cannot have two overlapping transpositions in the same Sylow $2$-subgroup, because their product is a $3$-cycle, e.g. $(12)(23) = (123)$, which has order $3$ and hence cannot be in any $2$-subgroup. This means that there are exactly two disjoint transpositions in each Sylow $2$-subgroup.


So, given the above, we know that the three Sylow $2$-subgroups must start like this:
$$S_1 = \{e, (12), (34), (12)(34),\ldots\}$$
$$S_2 = \{e, (13), (24), (13)(24),\ldots\}$$
$$S_3 = \{e, (14), (23), (14)(23),\ldots\}$$
Now reason similarly about the $4$-cycles.
A: You seem to have misunderstood some part of the Sylow theorems: It is not true that $n_2$ and $n_3$ divide $3$. The Sylow theorems state that $n_2\mid3$ and $n_2\equiv1\pmod2$, and that $n_3\mid2^3$ and $n_3\equiv1\pmod3$.
The problem is to find (and describe) two Sylow-subgroups of $S_4$. As you have noted, because $|S_4|=2^3\times3$ every Sylow-$3$ subgroup is of order $3$. It should not be difficult to find and describe such a subgroup.
Every Sylow-$2$ subgroup is of order $2^3=8$. So all its elements have order $1$, $2$, $4$ or $8$. There are no elements of order $8$ in $S_4$, so we should look at elements of orders $2$ and $4$. There aren't that many of those, and trying a few combinations should give you a subgroup of order $8$ rather quickly.
