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
$$P= \lbrace e, (1234), (13)(24), (1432),(24) ,(14)(23), (13), (12)(34)\rbrace.$$