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Examples and Applications

Example Using the Sylow Theorems, we can determine that A_5 has subgroups of orders 2, 3, 4, and 5. The Sylow p-subgroups of A_5 have orders 3, 4, and 5. The Third Sylow Theorem tells us exactly how many Sylow

Examples and Applications

Example

Using the Sylow Theorems, we can determine that \(A_5\) has subgroups of orders \(2\), \(3\), \(4\), and \(5\). The Sylow \(p\)-subgroups of \(A_5\) have orders \(3\), \(4\), and \(5\). 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 \(1 \pmod{5}\), there are either one or six Sylow \(5\)-subgroups in \(A_5\). 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 \(A_5\). Since \(A_5\) has no normal subgroups, this is impossible; hence, we have determined that there are exactly six distinct Sylow \(5\)-subgroups of \(A_5\).

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

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

Proof

We know that \(G\) contains a subgroup \(H\) of order \(q\). The number of conjugates of \(H\) divides \(pq\) and is equal to \(1 + kq\) for \(k = 0, 1, \ldots\). 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 \(G\).

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

Example

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

Example

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 \(9\), for some \(k = 0, 1, 2, \ldots\). Also, \(1 + 3k\) must divide \(11\); hence, there can only be a single normal Sylow \(3\)-subgroup \(H\) in \(G\). Similarly, there are \(1 +11k\) Sylow \(11\)-subgroups and \(1 +11k\) must divide \(9\). Consequently, there is only one Sylow \(11\)-subgroup \(K\) in \(G\). By , 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 \({\mathbb Z}_9\). Since \(K\) has order \(11\), it must be isomorphic to \({\mathbb Z}_{11}\). 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 \(5 \cdot 7 \cdot 47 = 1645\), we need the following theorem.

The subgroup \(G'\) of \(G\) is called the commutator subgroup of \(G\). We leave the proof of this theorem as an exercise ( in ).

Example

We will now show that every group of order \(5 \cdot 7 \cdot 47 = 1645\) is abelian, and cyclic by . By the Third Sylow Theorem, \(G\) has only one subgroup \(H_1\) of order \(47\). So \(G/H_1\) has order \(35\) and must be abelian by . Hence, the commutator subgroup of \(G\) is contained in \(H\) which tells us that \(|G'|\) is either \(1\) or \(47\). If \(|G'|=1\), we are done. Suppose that \(|G'|=47\). 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 \(G\), where \(|H_2| = 5\) and \(|H_3| = 7\). In either case the quotient group is abelian; hence, \(G'\) must be a subgroup of \(H_i\), \(i= 1, 2\). 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.

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 \(A_5\), 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

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.

Example

Let \(G\) be a finite group of order \(p^n\), \(n \gt 1\) and \(p\) prime. By , \(G\) 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 .

For other groups \(G\), 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.

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.

Sage will compute a single Sylow \(p\)-subgroup for each prime divisor \(p\) of the order of the group. Then, with conjugacy, all of the Sylow \(p\)-subgroups can be enumerated. It is also possible to compute the normalizer of a subgroup.

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

Symbols used here

a \equiv b \pmod n

congruent modulo n
n divides a − b; a and b have the same remainder.

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

imaginary unit
i² = −1.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

\blacksquare\ \text{or}\ \square

end of proof (halmos)
Marks the point where the statement has been established.

a \mid b,\ \gcd(a,b)

divides, greatest common divisor
b is a multiple of a; the largest number dividing both.

(G, \cdot),\ e,\ g^{-1}

group, identity, inverse
A set with an operation; the do-nothing element; the element that undoes g.

G \cong H,\ G / N

isomorphic, quotient group
Same structure; the group of cosets of a normal subgroup N.

\mathbb{Z}/n\mathbb{Z},\ \mathbb{Z}_n

integers modulo n
The remainders 0…n−1 with clock arithmetic.

\operatorname{Hom}(A, B),\ f \circ g

arrows from A to B, composition
The set of morphisms; do g then f.

Questions people ask

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group — that is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Wárá

Parts of this page are adapted from Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Condensed and re-explained here; errors are ours.

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