11.5.1 The Statements
By Lagrange’s theorem, the order of a subgroup of divides . The converse fails: a divisor of need not be the order of any subgroup.
Example 11.5.1 (Missing subgroups). A subgroup of index 2 contains every element of odd order (Lemma 11.2.6). So a subgroup of order 6 in would contain the identity and the eight 3-cycles, and a subgroup of order 30 in would contain the identity, the twenty 3-cycles and the twenty-four 5-cycles. Both are impossible: has no subgroup of order 6, and has no subgroup of order 30.
The Sylow theorems show that the converse does hold for prime powers, and they say much more.
Definition 11.5.2 (Sylow -subgroup). Let be a finite group and a prime, and write with . A Sylow -subgroup of is a subgroup of order . A -subgroup of is a subgroup that is a -group.
Theorem 11.5.3 (The Sylow Theorems (Sylow, 1872)). Let be a finite group and a prime, and write with .
- has a Sylow -subgroup.
- Any two Sylow -subgroups and of are conjugate: for some . More generally, every -subgroup of is contained in a Sylow -subgroup.
- The number of Sylow -subgroups of satisfies and .
We prove these theorems in Section 11.5.3. When they are trivial (, and is the only Sylow -subgroup), so the interesting case is . A few remarks:
- Since , is prime to , so is equivalent to .
- A conjugate of a Sylow -subgroup is again a Sylow -subgroup. So by (2), a Sylow -subgroup is normal if and only if it is the only one, that is, : if then for all , and if is normal then every Sylow -subgroup, being conjugate to , equals .
Corollary 11.5.4 (Cauchy’s theorem). If is a prime dividing , then contains an element of order .
Proof. By Sylow’s first theorem, has a subgroup of order . Pick , . The order of divides , so it is for some . Then has order . ∎
Note. For a -group the Sylow theorems say nothing: the only Sylow -subgroup is itself. In the Sylow approach to classifying finite groups the -groups are therefore the hardest case, and their number grows very fast with the exponent. For :
| order | | | | | |
|---|
| number of groups | 1 | 2 | 5 | 14 | 51 |
|---|
(order 4 is Corollary 11.2.5 and order 8 is Proposition 11.2.7), and there are already 56092 groups of order .
11.5.2 Normalizers and Conjugate Subgroups
A group acts by conjugation on the set of its subgroups, . The orbit of is the set of subgroups conjugate to , and is normal exactly when this orbit is .
Definition 11.5.5 (Normalizer). The normalizer of a subgroup is the stabilizer of for the conjugation action on subgroups. We write when is clear.
As a stabilizer, is a subgroup of , and it contains . For a subgroup with , is normal in if and only if for all , that is, if and only if . So is the largest subgroup of in which is normal, and if and only if . The normalizer measures how far is from being normal.
Example 11.5.6 (Normalizers in ). Let and . The subgroup is normal, so , even though does not commute with : . Conjugation by moves the elements of but preserves the subgroup. On the other hand, for we have , which equals if and only if commutes with ; the centralizer of has elements, so .
Proposition 11.5.7 (Counting conjugate subgroups). The number of subgroups of conjugate to is .
Proof. These subgroups form the orbit of for the conjugation action, whose stabilizer is . Apply the Orbit–Stabilizer Theorem. ∎
11.5.3 Proofs of the Sylow Theorems
Fix a finite group and a prime , and write with . The first two theorems are proved by letting act on suitable sets by left multiplication, and the third by letting a Sylow subgroup act by conjugation.
Lemma 11.5.8 (A binomial coefficient). If with , then .
Proof. We have The factor for is , which is prime to . For , write with ; then , and where neither nor is divisible by , since both are congruent to modulo . So the numerator and the denominator of each factor are divisible by exactly the same power of . Hence the power of in the numerator of the product equals the power of in its denominator, and the integer is not divisible by . ∎
Lemma 11.5.9 (Stabilizers of subsets). Let act on the set of its subsets by left multiplication, . For every subset , divides .
Proof. Let . Then for all and , so acts on by left multiplication. The orbit of is the right coset , which has elements. Since is the disjoint union of these orbits, divides . ∎
Proof of the First Sylow Theorem. Let be the set of all subsets of with elements, so . Left multiplication preserves the number of elements of a subset, so acts on by . By Lemma 11.5.8, . Since is the sum of the sizes of the orbits, some orbit has . Now is divisible by , and , so divides . On the other hand divides by Lemma 11.5.9. So , and is a Sylow -subgroup. (In fact is then a single orbit of , that is, a right coset of .) ∎
Proof of the Second Sylow Theorem. Let be a Sylow -subgroup, which exists by the first theorem, and let be a -subgroup of . Let act on the set of left cosets of by left multiplication; is not divisible by . Restrict this action to . The size of each -orbit divides , so it is a power of . Since and is the sum of the sizes of the -orbits, some orbit has size : there is a coset with for all . Then , that is, , for every . So , which is a Sylow -subgroup, since it has the same order as . If is itself a Sylow -subgroup, then and . ∎
Proof of the Third Sylow Theorem. Let act by conjugation on the set of its Sylow -subgroups. By the second theorem this action is transitive, and the stabilizer of a Sylow -subgroup is , so (Proposition 11.5.7). Since , for some integer , and divides .
Now restrict the conjugation action to . The orbit of itself is . We claim that it is the only orbit of size 1. Suppose is a Sylow -subgroup with for all , that is, . The order of is a multiple of and divides , so is the largest power of dividing , and and are Sylow -subgroups of . By the second theorem applied to the group , for some . But by the definition of , so . This proves the claim.
Every orbit of the -action has size dividing , so every orbit other than has size divisible by . Since is the sum of the sizes of the orbits, . ∎