Existence of Sylow subgroups; the first Sylow Theorem

Definition 10.1.1   Let be a finite group with and let be a prime such that but no higher power of divides . A subgroup of of order is called a -Sylow subgroup.

It is not at all obvious that a -Sylow subgroup exists. It is our main concern in this section to show that for each that a -Sylow subgroup exists. Note that is a -Sylow subgroup of if and only if where and .

We first consider and prove a very special case of the end result we wish to obtain.

Theorem 10.1.2   Let be a finite abelian group and let be a prime such that . Then contains at least one element of order .

Proof: The proof will proceed by induction on . If , a prime, then the theorem is clearly true. Thus suppose , where is composite and also suppose that the theorem has been proven for all groups whose order . Suppose is a prime such that . We need to show that has an element of order . We claim contains a subgroup not equal to or to itself. This is clear if is cyclic (see Theorem 5.2.1). If is not cyclic, let , , the identity element of . Then is a proper subgroup of . (Why?) Let be a proper subgroup of of maximal order. If , then since , we have by the induction hypothesis that there exists an such that , but clearly , also. This proves our claim when . If, however, , then since is a proper subgroup of , there exists an element . Let be the cyclic subgroup of generated by , i.e., . Now the product is a subgroup of (by Theorem 4.2.1) since is abelian. Also, properly, i.e., , because but . However, was a maximal proper subgroup. Thus it must be that . Then, by Theorem 4.3.6,

where . Thus

and since , we must have that . However, we have assumed , thus by the Corollary 1.2.10 we have . Let . Then , where , and consider the element . By Theorem 5.2.2,

Thus if is an abelian group and if , then contains a subgroup of order ; viz., the cyclic subgroup of order generated by an element of order whose existence is guaranteed by Theorem 10.1.2.

We now proceed to the main result of this section, i.e., the first Sylow Theorem.

Theorem 10.1.3 (Sylow I)   Let be a finite group and let , then contains a -Sylow subgroup (i.e., a -Sylow subgroup exists).

Proof: As in the preceding theorem, the proof will be given by induction on . The theorem is clearly true if . Now let , where . By hypothesis, . We next decompose into conjugacy classes according to the discussion at the beginning of Section 4.1 and use the class equation, i.e., equation (4.8), to obtain

 (10.1)

where designates a conjugacy class. Since the union is disjoint, we can write

where . By Theorem 4.3.4, , where . We return now to equation (10.1) and recall that the conjugacy classes, , listed (if there are any) are nontrivial, i.e., each . Let us suppose that some is such that , i.e., . Since , we must have . Moreover, since . It follows by the induction hypothesis that the subgroup contains a -Sylow subgroup and that therefore contains a -Sylow subgroup. Thus, in this case, the theorem has been established.

We may, thus, assume that for each , , . Thus

whence . Since is an abelian group, we have by the preceding theorem that , and therefore, has an element, , of order . Now since and . Hence , and so by the induction hypothesis must contain a -Sylow subgroup of order . This -Sylow subgroup must be of the form , where which contains by the Corollary 8.3.3. Now

and so is a -Sylow subgroup of .

On the basis of this theorem, we can now strengthen the result obtained in Theorem 10.1.2.

Theorem 10.1.4 (Cauchy)   If is a finite group and if is a prime such that , then contains at least one element of order .

Proof: Let be a -Sylow subgroup of , and let . If , then implies , where . But then the cyclic group, , must have a (unique) subgroup of order , say , by Theorem 5.2.1. Thus and .

Subsections

David Joyner 2007-08-06