§ Sylow Theorem 1
I've always wanted a proof I can remember, and I think I've found one.
- Let be a group such that where does not divide .
- We start by considering the set of all subsets of of size . Call this set .
- We will prove the existence of a special subset such that , and . That is, and . This is somewhat natural, since the only way to get subgroups out of actions is to consider stabilizers.
- We need to show the existence of an such that has maximal cardinality.
§ Lemma: :
- this is the coefficient of in . But modulo , this is the same as the coefficient of in . The latter is . Thus, (modulo ).
§ Continuing: Size of modulo :
- Let us begin by considering . This is since we pick all subsets of size from m. See that if we want to calculate , this is the coefficient of in . But modulo , this is the same as the coefficient of in . The latter is . Thus, (modulo ). Iterating the lemma shows us that . Thus, does not divide , since was the -free part of .
- This implies that there is some orbit whose size is not divisible by . --- Break into orbits. Since the left hand side is not divisible by , there is some term in the orbits size that is not divisible by .
- Let the orbit be generated by a set . So . Now orbit stabilizer tells us that . Since is not divisible by , this means that must be of size at least . It could also have some divisors of inside it.
- Next, we will show that can be at most .
§ Lemma: size of stabilizer of subset when action is free:
- Let a group act freely on a set . This means that for all group elements , if for any we have , then we must have . In logic, this is: .
- See that an implication of this is that for any two elements , we can have at most one such that . Suppose that we have two elements, such that and . This means that . But we know that in such a case, or .
- What does this mean? it means that for all .
- Now let's upgrade this to subsets of . Let (for part) be a subset of . What is ? We want to show that it is at most . Let's pick a unique basepoint [thus since ].
- Let's suppose that . This means that . Say it sends to . Now no other element of can send to since the action is free!
- Thus, there are at most choices for to be sent to, one for each element of .
- Thus, .
§ Continuing: Showing that .
- Since the action of on is free, and since we are considering the stabilizer of some subset , we must have that . Thus, since (from the orbit argument above) and (from the stabilizer argument), we have . Thus we are done.
- More explicitly perhaps, let us analyze . We know that . Thus, for any , we know that . Thus, .
Combining the above, we find that . So the stabilizer of
size it is in some sense "maximal": it has the largest size a
stabilizer could have!
- Also notice that is a coset of . Thus, .