# 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.

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

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

• The question does not ask you to find how many there are, it just asks you to describe one Sylow 2-subgroup and one Sylow 3-subgroup. The 3-subgroup is very easy; you just need to find an element of order 3. – Matt Samuel Dec 7 '15 at 3:05
• That's a good start. Finding a subgroup of order $8$ is a bit more work, but I'm sure you'll get there with some patience. Let us know when you've solved the problem (perhaps post your own answer?), or when you need some help. – Servaes Dec 7 '15 at 3:29
• Don't look for subsets to check whether they are subgroups. Look for generators of such a subgroup; pick a few elements of order $2$ and/or $4$, preferably as few as possible, and try to figure out how big the group is they generate. – Servaes Dec 7 '15 at 3:42
• @Servaes just a thought. Would the dihedral subgroup $D_{4}$ suffice since it is a subgroup of $S_{4}$ of order 8? – Jack Dec 7 '15 at 4:19
• @Tim: Nice observation! $D_4$ is indeed a subgroup of $S_4$, and since it has the correct order $8$, it is a Sylow $2$-subgroup. Indeed, since all Sylow $2$-subgroups are isomorphic (because they are conjugate), every Sylow $2$-subgroup is an instance of $D_4$ with different choices of labels for the vertices of the square. – Bungo Dec 7 '15 at 4:40

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.

• Care to explain the downvote? – Servaes Dec 7 '15 at 3:20

A few tips for finding a Sylow $2$-subgroup without too much manual labor:

1. 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.
2. 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 $4$.
3. 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.
4. 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.