If a positive integer divides the order of a finite group , does have a subgroup of order ? The answer is true for finite abelian groups, but it is not true for arbitrary groups. Sylow theorems discuss this situation when is a prime power.
Before discussing Sylow theorems, we first discuss group actions.
Let is a group and , the action of group on the set where is the product on is called a left translation. The action of on where is called conjugation by and the element is said to be a conjugate of . If is any subgroup of and , . Thus acts on the set of all subgroups of by conjugation . The group is said to be conjugate to .
The equivalence classes are called the orbits of on , denoted by for . The group is called the stabilizer of . If acts on itself by conjugation, the orbits are called conjugacy classes. If a subgroup acts on by conjugation, is called the centralizer of in and is denoted . is simply called the centralizer of . If acts by conjugation on the set of subgroups of , the subgroup of fixing , is called the normalizer of and is denoted . The group is simply the normalizer of . Every subgroup is normal in and is normal iff .
Proof. Let . Since it follows that is a well-defined bijection of the set of cosets of in onto . Hence . ◻
Proof. (i) and (iii) follow from the previous theorem and Lagrange’s theorem. Since conjugacy is an equivalence relation, (ii) follows from (i). ◻
Proof. If , define by . Since , is surjective. Similarly, implies whence is injective. Since , the map given by is a homomorphism. ◻
Proof. Let act on itself by left translation and obtain . If , then . In particular, whence and is a monomorphism. Note if , . ◻
If is a group, , the set of all automorphisms of is a group under composition.
Proof. (i) If acts on itself by conjugation, given by is a bijection. is also a homomorphism and hence an automorphism. (ii) Let act on itself by conjugation. The homomorphism has image contained in . Clearly whence . ◻
The automorphism is called the inner automorphism induced by . is called the center of . An element iff the conjugacy class of consists of alone. Thus if , then . Thus if is finite, then where are distinct conjugacy classes of and each . The above equation is called the class equation.
Proof. The induced homomorphism is given by where and . If and . In particular, implying . ◻
Proof. Apply the proposition. The kernel of the induced homomorphism is a normal subgroup of contained in and thus must be . Hence is a monomorphism. ◻
Proof. Let be the set of all left cosets of in . Then . If is the kernel of the homomorphism , and . Furthermore, is isomorphic to a subgroup of . Hence divides . But every divisor of must divide . Thus or . However, . Thus and , whence . But is normal in . ◻
We now discuss some lemmas that lead to the Sylow theorems.
Proof. An orbit contains exactly one element iff . Hence is a disjoint union with for all . Hence . for each since and divides . Therefore . ◻
Proof. Let be the -tuple of group elements . Since necessarily, , where . Since , . Let act on by cyclic permutations. . Note implies so that . Verify that for and . Thus the action is well-defined. Now iff . Clearly so . . There exists such that and hence . Since is prime, . ◻
In particular, is always a -subgroup of for every prime .
Proof. Consider the class equation . Since each and divides , and thus . ◻
Proof. Let be the set of left cosets of in and let act on by left translation. Then . Also, Thus . Then . ◻
Proof. . Since , we must have . Thus . ◻
Proof. Since , contains an element of order .Proceeding by induction, assume where for . Then and and . Hence and contains a subgroup of order . This group is of the form where is a subgroup of containing . Since is normal in , is necessarily normal in . Finally, . ◻
Sylow -subgroups always exist, though sometimes they may be trivial, and every -subgroup is contained in a Sylow -subgroup. The first Sylow theorem shows that a finite group has a nontrivial Sylow -subgroup for every prime that divides the order of .
Proof. Let be the set of left cosets of in and let act on by left translation. . But . Thus and there exists . If is a Sylow -subgroup, and hence . ◻
Proof. By the second Sylow theorem, the number of Sylow -subgroups is the number of conjugates of any one of them, say . This number is , a divisor of . Let be the set of all Sylow -subgroups of and let act on by conjugation. Then . Both and are Sylow -subgroups of and hence of and are therefore conjugate in . But meaning . Thus and . ◻
Proof. ◻