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.
(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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