## Section14.2Examples and Applications

###### Example14.9.

Using the Sylow Theorems, we can determine that $A_5$ has subgroups of orders $2\text{,}$ $3\text{,}$ $4\text{,}$ and $5\text{.}$ The Sylow $p$-subgroups of $A_5$ have orders $3\text{,}$ $4\text{,}$ and $5\text{.}$ The Third Sylow Theorem tells us exactly how many Sylow $p$-subgroups $A_5$ has. Since the number of Sylow $5$-subgroups must divide $60$ and also be congruent to $1 \pmod{5}\text{,}$ there are either one or six Sylow $5$-subgroups in $A_5\text{.}$ All Sylow $5$-subgroups are conjugate. If there were only a single Sylow $5$-subgroup, it would be conjugate to itself; that is, it would be a normal subgroup of $A_5\text{.}$ Since $A_5$ has no normal subgroups, this is impossible; hence, we have determined that there are exactly six distinct Sylow $5$-subgroups of $A_5\text{.}$

The Sylow Theorems allow us to prove many useful results about finite groups. By using them, we can often conclude a great deal about groups of a particular order if certain hypotheses are satisfied.

###### Proof.

We know that $G$ contains a subgroup $H$ of order $q\text{.}$ The number of conjugates of $H$ divides $pq$ and is equal to $1 + kq$ for $k = 0, 1, \ldots\text{.}$ However, $1 + q$ is already too large to divide the order of the group; hence, $H$ can only be conjugate to itself. That is, $H$ must be normal in $G\text{.}$

The group $G$ also has a Sylow $p$-subgroup, say $K\text{.}$ The number of conjugates of $K$ must divide $q$ and be equal to $1 + kp$ for $k = 0, 1, \ldots\text{.}$ Since $q$ is prime, either $1 + kp = q$ or $1 + kp = 1\text{.}$ If $1 + kp = 1\text{,}$ then $K$ is normal in $G\text{.}$ In this case, we can easily show that $G$ satisfies the criteria, given in Chapter 4, for the internal direct product of $H$ and $K\text{.}$ Since $H$ is isomorphic to ${\mathbb Z}_q$ and $K$ is isomorphic to ${\mathbb Z}_p\text{,}$ $G \cong {\mathbb Z}_p \times {\mathbb Z}_q \cong {\mathbb Z}_{pq}$ by Theorem 11.35.

###### Example14.11.

Every group of order $15$ is cyclic. This is true because $15 = 5 \cdot 3$ and $5 \not\equiv 1 \pmod{3}\text{.}$

###### Example14.12.

Let us classify all of the groups of order $99 = 3^2 \cdot 11$ up to isomorphism. First we will show that every group $G$ of order $99$ is abelian. By the Third Sylow Theorem, there are $1 + 3k$ Sylow $3$-subgroups, each of order $9\text{,}$ for some $k = 0, 1, 2, \ldots\text{.}$ Also, $1 + 3k$ must divide $11\text{;}$ hence, there can only be a single normal Sylow $3$-subgroup $H$ in $G\text{.}$ Similarly, there are $1 +11k$ Sylow $11$-subgroups and $1 +11k$ must divide $9\text{.}$ Consequently, there is only one Sylow $11$-subgroup $K$ in $G\text{.}$ By Corollary 13.16, any group of order $p^2$ is abelian for $p$ prime; hence, $H$ is isomorphic either to ${\mathbb Z}_3 \times {\mathbb Z}_3$ or to ${\mathbb Z}_9\text{.}$ Since $K$ has order $11\text{,}$ it must be isomorphic to ${\mathbb Z}_{11}\text{.}$ Therefore, the only possible groups of order $99$ are ${\mathbb Z}_3 \times {\mathbb Z}_3 \times {\mathbb Z}_{11}$ or ${\mathbb Z}_9 \times {\mathbb Z}_{11}$ up to isomorphism.

To determine all of the groups of order $5 \cdot 7 \cdot 47 = 1645\text{,}$ we need the following theorem.

The subgroup $G'$ of $G$ is called the commutator subgroup of $G\text{.}$ We leave the proof of this theorem as an exercise (Exercise 3.3.31).

###### Example14.14.

We will now show that every group of order $5 \cdot 7 \cdot 47 = 1645$ is abelian, and cyclic by Theorem 11.35. By the Third Sylow Theorem, $G$ has only one subgroup $H_1$ of order $47\text{.}$ So $G/H_1$ has order 35 and must be abelian by Theorem 14.10. Hence, the commutator subgroup of $G$ is contained in $H$ which tells us that $|G'|$ is either $1$ or $47\text{.}$ If $|G'|=1\text{,}$ we are done. Suppose that $|G'|=47\text{.}$ The Third Sylow Theorem tells us that $G$ has only one subgroup of order $5$ and one subgroup of order $7\text{.}$ So there exist normal subgroups $H_2$ and $H_3$ in $G\text{,}$ where $|H_2| = 5$ and $|H_3| = 7\text{.}$ In either case the quotient group is abelian; hence, $G'$ must be a subgroup of $H_i\text{,}$ $i= 1, 2\text{.}$ Therefore, the order of $G'$ is $1\text{,}$ $5\text{,}$ or $7\text{.}$ However, we already have determined that $|G'| =1$ or $47\text{.}$ So the commutator subgroup of $G$ is trivial, and consequently $G$ is abelian.

### Subsection14.2.1Finite Simple Groups

Given a finite group, one can ask whether or not that group has any normal subgroups. Recall that a simple group is one with no proper nontrivial normal subgroups. As in the case of $A_5\text{,}$ proving a group to be simple can be a very difficult task; however, the Sylow Theorems are useful tools for proving that a group is not simple. Usually, some sort of counting argument is involved.

###### Example14.15.

Let us show that no group $G$ of order $20$ can be simple. By the Third Sylow Theorem, $G$ contains one or more Sylow $5$-subgroups. The number of such subgroups is congruent to $1 \pmod{5}$ and must also divide $20\text{.}$ The only possible such number is $1\text{.}$ Since there is only a single Sylow $5$-subgroup and all Sylow $5$-subgroups are conjugate, this subgroup must be normal.

###### Example14.16.

Let $G$ be a finite group of order $p^n\text{,}$ $n \gt 1$ and $p$ prime. By Theorem 13.15, $G$ has a nontrivial center. Since the center of any group $G$ is a normal subgroup, $G$ cannot be a simple group. Therefore, groups of orders $4\text{,}$ $8\text{,}$ $9\text{,}$ $16\text{,}$ $25\text{,}$ $27\text{,}$ $32\text{,}$ $49\text{,}$ $64\text{,}$ and $81$ are not simple. In fact, the groups of order $4\text{,}$ $9\text{,}$ $25\text{,}$ and $49$ are abelian by Corollary 13.16.

###### Example14.17.

No group of order $56= 2^3 \cdot 7$ is simple. We have seen that if we can show that there is only one Sylow $p$-subgroup for some prime $p$ dividing 56, then this must be a normal subgroup and we are done. By the Third Sylow Theorem, there are either one or eight Sylow $7$-subgroups. If there is only a single Sylow $7$-subgroup, then it must be normal.

On the other hand, suppose that there are eight Sylow $7$-subgroups. Then each of these subgroups must be cyclic; hence, the intersection of any two of these subgroups contains only the identity of the group. This leaves $8 \cdot 6 = 48$ distinct elements in the group, each of order $7\text{.}$ Now let us count Sylow $2$-subgroups. There are either one or seven Sylow $2$-subgroups. Any element of a Sylow $2$-subgroup other than the identity must have as its order a power of $2\text{;}$ and therefore cannot be one of the $48$ elements of order $7$ in the Sylow $7$-subgroups. Since a Sylow $2$-subgroup has order $8\text{,}$ there is only enough room for a single Sylow $2$-subgroup in a group of order $56\text{.}$ If there is only one Sylow $2$-subgroup, it must be normal.

For other groups $G\text{,}$ it is more difficult to prove that $G$ is not simple. Suppose $G$ has order $48\text{.}$ In this case the technique that we employed in the last example will not work. We need the following lemma to prove that no group of order $48$ is simple.

###### Proof.

Recall that

\begin{equation*} HK = \{ hk : h \in H, k \in K \}\text{.} \end{equation*}

Certainly, $|HK| \leq |H| \cdot |K|$ since some element in $HK$ could be written as the product of different elements in $H$ and $K\text{.}$ It is quite possible that $h_1 k_1 = h_2 k_2$ for $h_1, h_2 \in H$ and $k_1, k_2 \in K\text{.}$ If this is the case, let

\begin{equation*} a = (h_1)^{-1} h_2 = k_1 (k_2)^{-1}\text{.} \end{equation*}

Notice that $a \in H \cap K\text{,}$ since $(h_1)^{-1} h_2$ is in $H$ and $k_2 (k_1)^{-1}$ is in $K\text{;}$ consequently,

\begin{align*} h_2 & = h_1 a^{-1}\\ k_2 & = a k_1\text{.} \end{align*}

Conversely, let $h = h_1 b^{-1}$ and $k = b k_1$ for $b \in H \cap K\text{.}$ Then $h k = h_1 k_1\text{,}$ where $h \in H$ and $k \in K\text{.}$ Hence, any element $hk \in HK$ can be written in the form $h_i k_i$ for $h_i \in H$ and $k_i \in K\text{,}$ as many times as there are elements in $H \cap K\text{;}$ that is, $|H \cap K|$ times. Therefore, $|HK| = (|H| \cdot |K|)/|H \cap K|\text{.}$

###### Example14.19.

To demonstrate that a group $G$ of order $48$ is not simple, we will show that $G$ contains either a normal subgroup of order $8$ or a normal subgroup of order $16\text{.}$ By the Third Sylow Theorem, $G$ has either one or three Sylow $2$-subgroups of order $16\text{.}$ If there is only one subgroup, then it must be a normal subgroup.

Suppose that the other case is true, and two of the three Sylow $2$-subgroups are $H$ and $K\text{.}$ We claim that $|H \cap K| = 8\text{.}$ If $|H \cap K| \leq 4\text{,}$ then by Lemma 14.18,

\begin{equation*} |HK| \geq \frac{16 \cdot 16}{4} =64\text{,} \end{equation*}

which is impossible. Notice that $H \cap K$ has index two in both of $H$ and $K\text{,}$ so is normal in both, and thus $H$ and $K$ are each in the normalizer of $H \cap K\text{.}$ Because $H$ is a subgroup of $N(H \cap K)$ and because $N(H \cap K)$ has strictly more than $16$ elements, $|N(H \cap K)|$ must be a multiple of $16$ greater than $1\text{,}$ as well as dividing $48\text{.}$ The only possibility is that $|N(H \cap K)|= 48\text{.}$ Hence, $N(H \cap K) = G\text{.}$

The following famous conjecture of Burnside was proved in a long and difficult paper by Feit and Thompson .

The proof of this theorem laid the groundwork for a program in the 1960s and 1970s that classified all finite simple groups. The success of this program is one of the outstanding achievements of modern mathematics.