p-Group and the Sylow Theorems

• (Fraleigh 36.1) Let be a group of order with prime power and let be a finite -set. Then

• Let be a prime. A group is a -group if , for some . A subgroup is a -subgroup if it is a subgroup of and also a -group.

• (Fraleigh 36.3) Cauchy’s Theorem. Let be a prime, and a finite group where divides . Then has an element of order and a subgroup of order .

  • Proof: Let be the set of all -tuples such that

    Now is divisible by since for each tuple, . Thus

    Consider the cycle acting on so that

    Note that (see Symmetric Group). The cyclic group has order . By (Fraleigh 36.1)

    The fixed points in correspond to tuples where . By the above congruence, there must be at least elements in . Thus, there must be some element , such that

• (Fraleigh 36.4) Let be a finite group. Then is a -group if and only if is a power of .

Sylow Theorems §

• We may view the Sylow Theorems as a counterpart to theorems about finite abelian groups.

• Let . The normalizer of in , denoted is defined as follows.

  That is, it is the set of all elements in that leave invariant under conjugation. It is also the largest subgroup having as a normal subgroup

• (Fraleigh 36.6) Let be a -subgroup of a finite group . Then

  See group indices.

• (Fraleigh 36.7) Let be a -subgroup of a finite group . If divides , then .

• (Fraleigh 36.8) First Sylow Theorem. Let be a finite group and let where and does not divide . Then

  • contains a subgroup of order for each .

  • Every subgroup of order is a normal subgroup of a subgroup of order .

  • Proof: Argue by induction. The base case follows from Cauchy’s Theorem.

    In the general case, suppose we have a subgroup of order where . . We have

    Since is the normalizer, . Therefore is well defined and divides . We can apply Cauchy’s theorem to the factor group to get a subgroup of order .

    Let be the canonical homomorphism. Then . here has order and it is clearly normal in .

• A Sylow -subgroup of a group is a maximal -subgroup of . That is, it is not contained in any larger -subgroup.

  • The Sylow -subgroups of a group with order is precisely the subgroups of order .
  • If is a Sylow -subgroup, every conjugate of is also a Sylow -subgroup.

• (Fraleigh e36.12) Let be a finite group and be primes that divide . If has precisely one proper Sylow -subgroup, it is a normal group so is not simple.

  • Proof: There is only one such subgroup . Therefore every conjugate of is a Sylow -subgroup, and thus for all . Thus, is normal.

• (Fraleigh 36.10) Second Sylow Theorem. Let and be Sylow -subgroups of a finite group . Then and are conjugate subgroups of .

  • Proof: Let be the collection of left cosets of and act on by the group action
    Where . Thus, is a -set. Therefore
    Therefore . Let then
    Which implies
    And since because they are both Sylow -subgroups, we have

• (Fraleigh 36.11) Third Sylow Theorem. If is a finite group and where does not divide , then the number of Sylow -subgroups, denoted satisfies the following properties

  • divides . In fact

  • where is any Sylow -subgroup.

  • Proof: Let be the set of all Sylow -subgroups with . Let act on by conjugation. We have . Note that .

    If then . Also and since both and are both Sylow -subgroups of , they are for as well so the Second Sylow Theorem implies they are conjugates.

    Since it is only conjugate in so . Thus is a stabilizer of and which proves the third statement. Also

    which proves the second statement.

    Also let act on by conjugation. There is only one orbit in under G by the Second Sylow Theorem since they are conjugates of each other. We therefore have for

    By the Orbit Stabilizer Theorem, it follows that which divides which proves the first statement.

• (Fraleigh e36.16) Let be a finite group and be a prime dividing . Let be a Sylow -subgroup. Then

  • Idea: From the Third Sylow Theorem, is a stabilizer of the group action on by .

• (Fraleigh 37.1) Every finite -group is solvable.

  • Idea: From the First Sylow Theorem, each -subgroup is normal in . All have order . By (Fraleigh 10.11) they are all cyclic and hence Abelian.

• (Fraleigh 37.4) The center of a finite nontrivial -group is nontrivial.

  • Proof: In the Classic class equation, divides . Since , there must exist

• (Fraleigh e36.21) Let be a prime. A finite group of order contains a normal series. where for .

  • Idea:The normal series in question is the ascending central series of

• (Fraleigh 37.5) Let such that and . Then

  See direct products

  • Idea: Show that for and using the commutator. Then construct a homomorphism defined by . What remains is to show that is an isomorphism.

• (Fraleigh 37.6) For a prime number , every group of order is abelian.

  • Proof: If is not cyclic, there must exist such that and .

    Clearly by the First Sylow Theorem.

    Also . Also since .

    Therefore (Fraleigh 37.5) can be applied to get which proves that it is Abelian by the Fundamental Theorem of Finitely Generated Abelian Groups.

• (Fraleigh 37.7) If and are distinct primes with , then every group of order has a single subgroup of order which is normal in . Hence is not simple.

  If then is Abelian and Cyclic.

• (Fraleigh 37.9) No group of order for is simple where is prime.

Links §

• A First Course in Abstract Algebra 7th Edition by Fraleigh