TITLE: In $S_{4}$, find a Sylow 2-subgroup and a Sylow 3-subgroup.
QUESTION [6 upvotes]: 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
$$P= \lbrace e, (1234), (13)(24), (1432),(24) ,(14)(23), (13), (12)(34)\rbrace.$$

REPLY [4 votes]: 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.