Existence of Sylow subgroups; the first Sylow Theorem

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.

**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.

**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

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.

**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 .

David Joyner 2007-08-06