An interesting concept from group theory is that of a Sylow-\(p\) subgroup.
Definition4.1.Sylow-\(p\) subgroup.
If \(G\) is a finite group, and \(p\) is a prime number, and \(p^k\) is the largest power of \(p\) dividing \(\order{G}\text{,}\) then a Sylow-\(p\) subgroup of \(G\) is a subgroup of \(G\) with order \(p^k\text{.}\)
This definition allows \(k\) to be \(0\text{,}\) so \(p\) might not divide \(\order{G}\text{.}\) There is a theorem that gives information about how many Sylow-\(p\) subgroups a finite group might have.
Theorem4.2.Sylow Theorem.
If \(G\) is a finite group and \(p\) is a prime dividing \(\order{G}\text{,}\) then the number of Sylow-\(p\) subgroups of \(G\) divides \(\order{G}\) and is congruent to \(1\) modulo \(p\text{.}\)
Example4.3.Sylow-\(3\) subgroups, group of order \(45\).
If \(G\) is a group of order \(45\text{,}\) how many Sylow-\(3\) subgroups could \(G\) have?
Write \(n_3\) to represent the number of Sylow-\(3\) subgroups. We factor \(45=9\cdot5\text{,}\) separating the maximal power of \(3\) from its complement. Since \(n_3\) divides \(45\) and \(n_3\equiv1\) modulo \(3\text{,}\) it follows that \(n_3\) divides \(5\text{.}\) So either \(n_3=1\) or \(n_3=5\text{.}\) But of those two options, only \(1\) is congruent to \(1\) mod \(3\text{.}\) So \(G\) must have \(1\) Sylow-\(3\) subgroup.
Checkpoint4.4.Sylow-\(7\) subgroups, group of order \(63\).
If \(G\) is a group of order \(63\text{,}\) how many Sylow-\(7\) subgroups could \(G\) have?
Write \(n_7\) to represent the number of Sylow-\(7\) subgroups. We factor \(63=9\cdot7\text{,}\) separating the maximal power of \(7\) from its complement. Since \(n_7\) divides \(63\) and \(n_7\equiv1\) modulo \(7\text{,}\) it follows that \(n_7\) divides \(9\text{.}\) So either \(n_7=1\text{,}\)\(n_7=3\text{,}\) or \(n_7=9\text{.}\) But of those, only \(1\) is a real possibility, since only \(1\) is congruent to \(1\) mod \(7\text{.}\) So \(G\) has \(1\) Sylow-\(7\) subgroup.
Checkpoint4.5.Sylow-\(3\) subgroups, group of order \(63\).
If \(G\) is a group of order \(63\text{,}\) how many Sylow-\(3\) subgroups could \(G\) have?
\(G\) either has \(1\) Sylow-\(3\) subgroup or it has \(7\text{.}\)
Solution.
Write \(n_3\) to represent the number of Sylow-\(3\) subgroups. We factor \(63=9\cdot7\text{,}\) separating the maximal power of \(3\) from its complement. Since \(n_3\) divides \(63\) and \(n_3\equiv1\) modulo \(3\text{,}\) it follows that \(n_3\) divides \(7\text{.}\) So either \(n_3=1\) or \(n_3=7\text{.}\) Both of those are real possibilities, since both are congruent to \(1\) mod \(3\text{.}\) So \(G\) either has \(1\) Sylow-\(3\) subgroup or it has \(7\text{.}\)
Understanding how to count Sylow-\(p\) subgroups will help us later when we are classifying finite groups.
Reading QuestionsReading Questions
1.
Does a group of order \(21\) have a Sylow-\(5\) subgroup?
2.
Does Theorem 4.2 always tell you exactly how many Sylow-\(p\) subgroups a finite group \(G\) has when you know the order of \(G\text{?}\)
ExercisesExercises
Vocabulary
1.
Research “Sylow” and explain who this person was with a short, one-paragraph biography.
2.
Given a group \(G\) of order \(72\text{,}\) what would be the order of a Sylow-\(2\) subgroup?
Calculations
3.
If \(G\) is a group of order \(2p\) for an odd prime \(p\text{,}\) how many Sylow-\(p\) subgroups might \(G\) have?
Exercise Group.
Let \(G\) be a group of order \(30\text{.}\) For each \(p\text{,}\) how many Sylow-\(p\) subgroups might \(G\) have?
4.
\(p=2\)
5.
\(p=3\)
6.
\(p=5\)
Thinking Deeper
7.
If \(G\) has only one Sylow-\(p\) subgroup for some prime \(p\text{,}\) prove that it is a normal subgroup.
8.
If \(\order{G}=pq\) for primes \(p\lt q\text{,}\) what conditions on \(p\) and \(q\) guarantee that \(G\cong C_p\times C_q\text{?}\)
