######
Example 15.9

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.

######
Theorem 15.10

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.

######
Example 15.11

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

######
Example 15.12

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.

######
Theorem 15.13

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

######
Example 15.14

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 \(47\text{.}\) 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.

###
Subsection Finite 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.

######
Example 15.15

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.

######
Example 15.16

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.

######
Example 15.17

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.

######
Lemma 15.18

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

######
Example 15.19

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| = \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 \(1\text{,}\) 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].

######
Theorem 15.20 Odd 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.