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Section15.2Examples and Applications

¶

Example15.9

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.

Theorem15.10

If $p$ and $q$ are distinct primes with $p \lt q \text{,}$ then every group $G$ of order $pq$ has a single subgroup of order $q$ and this subgroup is normal in $G\text{.}$ Hence, $G$ cannot be simple. Furthermore, if $q \not\equiv 1 \pmod{p}\text{,}$ then $G$ is cyclic.

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 9, 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 9.21.

Example15.11

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

Example15.12

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; 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 14.16, 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, 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.

Theorem15.13

Let $G' = \langle a b a^{-1} b^{-1} : a, b \in G \rangle$ be the subgroup consisting of all finite products of elements of the form $aba^{-1}b^{-1}$ in a group $G\text{.}$ Then $G'$ is a normal subgroup of $G$ and $G/G'$ is abelian.

The subgroup $G'$ of $G$ is called the of $G\text{.}$ We leave the proof of this theorem as an exercise (Exercise 10.3.14 in Chapter 10).

Example15.14

We will now show that every group of order $5 \cdot 7 \cdot 47 = 1645$ is abelian, and cyclic by Corollary 9.21. 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 15.10. Hence, the commutator subgroup of $G$ is contained in $H$ which tells us that $|G'|$ is either 1 or 47. 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. 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, 5, or 7. However, we already have determined that $|G'| =1$ or 47. So the commutator subgroup of $G$ is trivial, and consequently $G$ is abelian.

SubsectionFinite 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.

Example15.15

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. The only possible such number is 1. Since there is only a single Sylow 5-subgroup and all Sylow 5-subgroups are conjugate, this subgroup must be normal.

Example15.16

Let $G$ be a finite group of order $p^n\text{,}$ $n \gt 1$ and $p$ prime. By Theorem 14.15, $G$ 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, 8, 9, 16, 25, 27, 32, 49, 64, and 81 are not simple. In fact, the groups of order 4, 9, 25, and 49 are abelian by Corollary 14.16.

Example15.17

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. 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; 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, there is only enough room for a single Sylow 2-subgroup in a group of order 56. 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. 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.

Lemma15.18

Let $H$ and $K$ be finite subgroups of a group $G\text{.}$ Then

$$\begin{equation*} |HK| = \frac{|H| \cdot |K|}{|H \cap K|}. \end{equation*}$$

Proof

Recall that

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

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

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

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,

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

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, $|HK| = (|H| \cdot |K|)/|H \cap K|\text{.}$

Example15.19

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. By the Third Sylow Theorem, $G$ has either one or three Sylow 2-subgroups of order 16. 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 15.18,

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

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, as well as dividing 48. 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 [2].

Theorem15.20Odd Order Theorem

Every finite simple group of nonprime order must be of even order.

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.