The * Sylow theorems* describe the p-subgroups of
a finite group, where p is a prime number.

* Definition.* Let G be a group, and let p be a
prime number. A * Sylow p-subgroup* is a maximal
p-subgroup of G.

To be specific, if H is a Sylow p-subgroup of G and K is a p-subgroup of G such that , then .

* Remark.* Every p-subgroup of G is contained
in a Sylow p-subgroup.

Suppose that G is finite, and let H be a p-subgroup. If H isn't properly contained in any other p-subgroup, it's a Sylow p-subgroup by definition. Otherwise, let be a p-subgroup properly containing H.

Continuing in this way, I get a properly increasing chain of p-subgroups

But G is finite, so this chain must be finite, and the process must stop with a Sylow p-subgroup.

If G is infinite, the result is still true, but requires the use of
* Zorn's lemma*. I won't prove it here, since I'm
only considering finite groups in this section.

* Lemma.* Let G be a group and let p be a prime
number. Suppose that H is a normal p-subgroup of G and is a
p-group. Then G is a p-group.

* Proof.* Let . I must show that the order
of g is a power of p.

Since is a p-group, the order of is a power of p. Suppose that , i.e. . Thus, .

Since H is a p-group, the order of is a power of p. Suppose that . Then , and the order of g must be a divisor of . Therefore, the order of g is a power of p, and G is a p-group.

* Lemma.* Let G be a group, and let p be a
prime number. G acts on the set of Sylow p-subgroups of G by
conjugation.

* Proof.* The result is vacuously true if there
are no Sylow p-subgroups. Otherwise, it suffices to show that that
conjugate of a Sylow p-subgroup is a Sylow p-subgroup, since after
that the axioms for an action are clearly satisfied.

Let H be a Sylow p-subgroup of G, and let . I must show that is a Sylow p-subgroup.

Since conjugation is an automorphism, it carries subgroups to subgroups. Therefore, is a subgroup.

If , then h has order for some . Since conjugation is an automorphism, it preserves the orders of elements. Therefore, also has order . It follows that is a p-subgroup.

Finally, I must show that is maximal. Suppose that K is a p-subgroup and . Then , and is a p-subgroup because conjugation by is an automorphism. Since H is maximal, Hence, .

This proves that is a Sylow p-subgroup, as I wished to show.

* Remarks.*

(a) The isotropy group of a Sylow p-subgroup H under the conjugation action is the normalizer .

(b) If G has only one Sylow p-subgroup, it must be normal.

* Lemma.* Let G be a group, let p be a prime
number, and let H be a Sylow p-subgroup. Then no nontrivial element
of has order which is a power of p.

* Proof.* Suppose on the contrary that has
order a power of p, where . I want to show that is
the identity.

If the order of is , then . Then any element of has order dividing , so is a p-subgroup of .

Subgroups of are in one-to-one correspondence with subgroups of which contain H. Thus, there is a subgroup K of such that .

Now H and are p-groups, so K is a p-group, by an earlier lemma. Since K is a p-group and , maximality of H implies that .

Thus, , which is what I wanted to show.

* Lemma.* Let G be a group, let p be a prime
number, let H be a Sylow p-subgroup, and let . If the order of
g is a power of p and , then .

* Proof.* If , then . Since the order of g is a power of p, the same is
true of the order of . By the preceding lemma, ,
and .

* Theorem.* (Sylow) Let G be a finite group,
and let p be a prime number.

(a) The number of Sylow p-subgroups is congruent to 1 mod p.

(b) Any two Sylow p-subgroups are conjugate.

(c) The number of Sylow p-subgroups divides .

* Proof.* Let H be a Sylow p-subgroup. Consider
the set of conjugates of H

By an earlier result, all the H's are Sylow p-subgroups.

G acts on S by conjugation. Since , H also acts on S by conjugation. Consider the orbits under this action of H. The order of an orbit is the index of a corresponding isotropy group, and . Since the order of H is a power of p, it follows that the order of each orbit is a power of p.

Thus, the order of an orbit is divisible by p --- unless the orbit has order . In this case, for some i and all . Applying the preceding lemma, I find that for all , i.e. . Since is a Sylow p-subgroup, this implies that .

Thus, *if* an orbit consists of a single element,
*then* it is . Conversely, if , then .

Thus, the orders of the orbits of S under conjugation by H are all divisible by p, except for . It follows that , which is the number of conjugates of H, is congruent to 1 mod p.

Next, let K be a Sylow p-subgroup, and suppose that K is *not*
conjugate to H. As above, since G acts of S by conjugation and ,
K acts on S by conjugation. Again arguing as above, I find that the
order of each orbit of S under conjugation by K is a power of p.

Could there be an orbit consisting of a single element? Again arguing as above, I conclude that if is a one-element orbit, then . But this is ruled out, because K was not a conjugate of H.

In this case, then, the orders of *all* the orbits of S under
conjugation by K are divisible by p. It follows that , which contradicts the fact that established earlier.

This proves that every Sylow p-subgroup is a conjugate of H.

Finally, the number of conjugates of H is equal to the index , which divides . Therefore, the number of Sylow p-subgroups divides .

* Corollary.* (Sylow) Let G be a group of order
, where p is prime and . Then every
Sylow p-subgroup of G has order .

* Proof.* Let H be a Sylow p-subgroup of G. I
have

Since is the number of Sylow p-subgroups, it's congruent to 1 mod p by the preceding theorem. Therefore, is not divisible by p.

I claim that is not divisible by p. Indeed, if , then by Cauchy's theorem has an element of order p. This contradicts an earlier lemma.

Since p does not divide or , it does not divide .

Now for some , so

Then , but . Therefore, , and . This shows that .

* Example.* Let G be a group of order 15. Show
that G is cyclic.

If H and K are subgroups of order 5, then is a subgroup of H, so . Hence, or . In the second case, the subgroups are identical. In the case where the subgroups are distinct, they intersect in the identity only, and each contains 4 elements which are not in the other subgroup.

The number of Sylow subgroups of order 5 is congruent to 1 mod 5, so it could be 1, 6, 11, 16, .... None of 6, 11, or 16 divides 15, and if there were more than 16 Sylow-5 subgroups, the preceding discussion shows that they would contain more elements than there are in the group!

Consequently, there is a single Sylow-5 subgroup. Since there is a unique subgroup of order 5, it must be normal.

Arguing in the same way as with the Sylow-5 subgroups, I find that two Sylow-3 subgroups are either identical, or intersect only in the identity. Consequently, two distinct Sylow-3 subgroups each contains 2 elements which are not in the other.

The number of Sylow subgroups of order 3 is congruent to 1 mod 3, so it could be 1, 4, 7, .... Now 4 and 7 don't divide 15, and if there are more than 7 Sylow-3 subgroups, the preceding argument shows that there are at least elements in these subgroups. But I already have 4 elements of order 5 in the Sylow-5 subgroup, so this is impossible.

Consequently, there is a single Sylow-3 subgroup. Since there is a unique subgroup of order 3, it must be normal.

The Sylow-3 and Sylow-5 subgroups account for all the elements of order 3 or order 5, and there are elements between them. This leaves elements unaccounted for. Since elements of G can have order 1, 3, 5, or 15, there must be elements of order 15, so G is cyclic.

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Copyright 2012 by Bruce Ikenaga