Sylow Groups

#matematik-groups

Definition: Conjugate

Let be a group.
is conjugate to if there exists a such that .
is conjugate to if there exists a such that .

Definition: -group, -Sylow subgroup

Let be a finite group and be prime.
is a -group if for some .
is called a -Sylow subgroup if for some and where .

In other words, given a group whose order is the prime factorization , a
-Sylow subgroup is a subgroup with order .

Lemma

Let be a group with for some prime and some , and let be the number of subgroups in of order . Then In particular, is dependent only on the order of .

Proof:
Let . Then acts on via . With respect to this group action, let be a system of representatives of the -orbits. Then by the Orbit stabilizer theorem and Orbit equation.

Claim 1

For every , is a -group.

Proof: We have , meaning acts on via . Since this action is injective, we know each orbit has size . This means we get meaning so for some .

From this claim, we now get that meaning

Claim 2

Proof: Each corresponds to a unique such that as -orbits in , since all 's are a system of representatives for the orbits. This defines a map

Claim a)

Proof: Let . Then, like before, there is an such that , as orbits in . From before, we have We also know so we know meaning , so .

Claim b)

Proof: Let , meaning . From before, we have that This means must be the only orbit, meaning we can choose a and get that . This means where the first equality holds because . Then, since , we can simply let making a subgroup of that satisfies , meaning , so .

Claim c) is injective

Proof: Let such that . Then we have , meaning there is a such that , meaning is a group, so meaning .

Hence is a bijection between and .

By definition, , so from earlier we have

Theorem

Let be a group with for some prime and some , and let be the number of subgroups in of order . Then In particular, .

Proof:
We know the cyclic group has a unique subgroup of order , meaning , meaning , and by Lemma, we thus know that the latter is true for too.

Lemma

Let be a group, be a -subgroup and be a -Sylow subgroup. Then there exists a such that .

Proof:
We let act on via . We then get that which is clearly not divisble by . We also know that every orbit has a representative for some , giving which we know divides for some , so for some , and since the sum of all sizes of orbits are not divisible by , some must be , so that for some , meaning that for every ,

Theorem: Sylow Theorems

Let be a finite group and be prime.

  1. contains at least one -Sylow subgroup. More precisely, for any -subgroup there is a -Sylow subgroup such that .
  2. Let be a -Sylow subgroup and . Then is a -Sylow subgroup if and only if is conjugate to .
  3. Let be the number of -Sylow subgroups of . Then and .

Proof:
4. {a} First part follows from Theorem, second from the first part and Lemma.
5. : Clear since conjugation is an automorphism.
: By Lemma there exists a such that meaning since they are of the same size.
6. Let . Then by b) we know acts transitively on , meaning only one orbit exists. Hence for every , , so From Theorem we also know .

Lemma

Let be a group and such that . Then

  1. For every and , .
  2. is a group homomorphism.
  3. is injective.

Proof:
4. {a} We have since is normal, and likewise, since , too, is normal. Since , we thus know .
5. We have by (a).
6. We have

Corollary

Let be a finite group and be prime.

  1. If , then there exists a such that .
  2. is a -group for every there exists a such that .
  3. Let . Then is a -Sylow subgroup of if and only if is a maximal -group in .
  4. Let be a -Sylow subgroup. Then .

Proof:
5. {a} We know by Theorem that there is a -Sylow subgroup of some order for some positive . This means there is an where , and we know , so for some positive . We then get that has order since .
6. Let . We have since .
: Let be a prime such that . Then by (a) there is a such that , which by our assumption means .
7. : We know and for some and where . Then we know any -subgroup has order where since its order must divide , meaning it is at most as large as .
: Since is a -group in , there is a -Sylow subgroup containing , and by the maximality of , must be that -Sylow subgroup.
8. By Theorem we know all -Sylow subgroups are conjugate, meaning if and only if there is only one -Sylow subgroup, it is invariant under conjugation, meaning it's normal.