望远镜无助于限定灵魂。
# The First Sylow Theorem
The first Sylow theorem can be viewed as a specific converse of the Lagrange theorem.
(Sylow I) Given a group such that
where is the largest power of (that is, ), then there exists a subgroup such that
Such a subgroup is called a Sylow -subgroup. It has maximal prime power order within .
Example: Consider , since , the Sylow theorem states that there is a subgroup of order .
In fact, we can take
Example: For is the dihedral group, we have . So there must be subgroups of size 5 and 2. A subgroup generated
by a rotation has order 5. A subgroup generated by any reflection has order 2.
(Corollary of Sylow I) If prime number divides , then there exists an element with order .
Proof: Using Sylow I, there exists a subgroup such that . Pick any element , one knows and thus the order of is by Lagrange Theorem. Then is an element of order .
# The Second Sylow Theorem
The first Sylow theorem states that a Sylow -subgroup, a subgroup of maximal size dividing , exists. In fact, we can say a lot more about what these subgroups look like.
(Sylow II) There are two parts; part one is what is usually referred to as the second Sylow theorem.
- Given , where is a Sylow -subgroup, any other Sylow -subgroup is conjugate to ; i.e., there exists such that .
- Given any subgroup such that , for any Sylow subgroup , there exists such that .
Notice that does not have to be the maximal prime power, and can have order smaller than . Every prime power order subgroup, up to conjugation, sits inside a Sylow subgroup.
Evidently, conjugating a Sylow subgroup will result in a Sylow subgroup (since they have the same size), and Sylow II states that all the Sylow subgroups arise in this way.
Note that the second part is stronger, since can be a prime power smaller than , and implies the first part by applying to .
The second part states that given any prime power subgroup and any Sylow subgroup , it is possible to conjugate to make it land in .
Since the conjugation of a group is also a group, with the same size, thus the converse of the first part is true.
Example: For , every subgroup of size 2 is generated by a refection, and Sylow II indicates that all the refections are conjugate.
# The Third Sylow Theorem
The last Sylow theorem indicates the number of these (conjugate) subgroups.
(Sylow III) The number of Sylow -subgroups of divides
and is congruent to 1 modulo .
This theorem seems kind of weird, but is actually very useful.
Example: Consider and . The number of Sylow −subgroups is , which does divide and is congruent
to modulo .
Summary: The first Sylow theorem indicates existence of Sylow subgroups, the second Sylow theorem indicates that all Sylow subgroups are related by conjugation, and the third provides strong (and kind of funky) constraints on the number of such subgroups.
# Applications of the Sylow Theorems
Example: Consider any group of order , by Sylow III, for , the number of Sylow -subgroups divides and is equal to 1 modulo 5. Therefore the only possibility is that the number of Sylow 5-subgroup is 1.
So there is a unique such that . Since there is only one subgroup of size 5, Sylow II indicates that .
That is, is normal: the Sylow theorems indicate automatically that there is a normal subgroup of size 5.
For , Sylow III states that the number of Sylow 3−subgroups divides 5 and is 1 mod 3, so it is also 1.
Thus, there exists some unique such that .Moreover, , since is the cyclic group of order 5 and is the cyclic group of order 3.
Nontrivial elements of have order 5, while elements of have order 3, so they intersect only at the identity.
So the Sylow theorems give, for free, nonintersecting normal subgroups of size 5 and 3 for every single group of size 15. That is quite impressive!Moreover, we can prove that
- for , ;
- .
Proof: For the first statement, since is normal, thus , and therefore .
Similarly, since is normal, . Because , thus and .
The second part follows from the first one, one can verify that is an isomorphism.
Another application of the Sylow theorems is the decomposition of finite abelian groups.
Consider an abelian group such that
Then we know that we have a Sylow subgroup such that
Since is abelian, conjugating a group produces the same group, so by Sylow II, these (abelian) subgroups are unique for each prime.
Every finite abelian group is isomorphic to a product of groups of prime power order.
One can verify that , where is an isomorphism.
# Proof of Sylow I
Let be the set of subsets of of size , and be the order of group . Then by basic combinatorics,
Let act on by left translations:
Our eventual goal is to fnd a subgroup of of size by looking at stabilizers, as they are always subgroups of .
We find the size of a stabilizer by trying to find an orbit of size (where ), as we then know that the stabilizer will be order by orbit-stabilizer theorem.
Lemma: For such that , one has
考虑展开 考察系数?这个就是一个数论问题。
Lemma: Suppose we have a subset , which is a subset of . Also, let be a subgroup of that stabilizes . Then, divides .
Proof: Since stabilizes , for any , . In other words, for any ,
This implies that the right cosets partition . Since the cosets have the same size, we know that divides .
We know that the orbits partition , so
Since does not divide LHS, there must exist an orbit such that
注意, 是 的子集,是一个 “大小为 的 的子集的集合”。
Let the size of be .
Now, consider some element . By the orbit-stabilizer theorem, we also know that
And so and since . By the second lemma, divides , thus .
# Proof of Sylow II
Fix to be a Sylow subgroup. Consider a set , the left cosets of . The index of is the same as , so .
Let be the subgroup we want to show is a subgroup of up to conjugation, where .
We will look at how acts on by left translation, the mapping:
We decompose into orbits, . Note that these orbits are with respect to the action of , not the action of , as that would be transitive and we’d only have one orbit. That is, for .
We have that divides (因为 orbit-stabilizer theorem, ), but does not divide . Thus this orbit decomposition can only work if some orbit has size 1.
In other words, there exists some coset that is fixed by all . Then,
for all . Thus .