## 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.

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 $$11\text{;}$$ 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 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 $$11\text{,}$$ 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 10.4.14 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 Theorem 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 $$47\text{.}$$ 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.

### 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 $$20\text{.}$$ 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.

#### 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 $$4\text{,}$$ $$8\text{,}$$ $$9\text{,}$$ $$16\text{,}$$ $$25\text{,}$$ $$27\text{,}$$ $$32\text{,}$$ $$49\text{,}$$ $$64\text{,}$$ and $$81$$ are not simple. In fact, the groups of order $$4\text{,}$$ $$9\text{,}$$ $$25\text{,}$$ and $$49$$ are abelian by Corollary 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 $$7\text{.}$$ 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.

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

#### 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 $$16\text{.}$$ 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 15.18,

\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 [2].

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.