Sylow’s theorems are direct applications of group actions and form a generalisation of Cauchy’s theorem.
The Existence Theorem
First Sylow's Theorem
Let be a group of order, where is a prime and . Then has a subgroup of order , we call it -Sylow subgroup.
Proof Without loss of generality, we can assume that . We use induction on . Clearly for there is nothing to prove. Assume it is true for all groups with order smaller than . Suppose it is false for , that is, has no subgroups of order . Then, by our induction hypothesis, for every proper subgroup , . In particular, if , then its centraliser is a proper subgroup of , and hence . By class equation, we have where are representatives of the non-trivial conjugacy classes. Since and , we must have . By Cauchy’s theorem, there exists an element of order . Let be the subgroup generated by . Then and consider the quotient group . We have By our induction hypothesis, there exists a subgroup of order in , say . Let be the quotient map. Then is a subgroup of order in (ref. correspondence theorem), which is a contradiction. Therefore, our assumption that has no subgroups of order is false.
e.g.
Consider the symmetric group, whose order is . Sylow’s I states that has a subgroup of order . Indeed, the subgroup is a subgroup of order .
For the dihedral group of order , Sylow’s I states that has a subgroup of order . Indeed, the subgroup generated by a rotation is a subgroup of order .
The Classification Theorem
Lemma
Suppose is a finite set and is a -group acting on , and does not divide , then the action must have a fixed point.
Proof Every orbit size is a power of (ref. Orbit-stabliser theorem), so each orbit has size either or a power of , and a sum of such numbers can be non-multiple of only if at least one orbit has size .
Any two -Sylow subgroups in a finite group are conjugate;
Any -subgroup of is contained in a -Sylow subgroup.
Proof Let be a -subgroup, and be a -Sylow subgroup. We will show that there is such that . Once we have that, both parts of the theorem are proved. Consider the (left multiplication) action of on the left coset space. While this is transitive, the (sub) action of must have a fixed point by the above lemma. Let be a fixed point of the action of , then for all , we have , which implies that . Therefore, .
Remark
The converse of the first part of Sylow II also holds. That is, If is a -Sylow subgroup, is also a -Sylow subgroup, because conjugating by an element evidently produces a subgroup of the same size.
e.g. For , every subgroup of size is generated by a refection, and Sylow II indicates that all the refections are conjugate.
Corollary
A group has exactly one -Sylow subgroup if and only if is normal.
The Counting Theorem
Third Sylow’s Theorem
The number of $p$-Sylow subgroups in a finite group of order with is of the form and divides .
Proof Let . Consider the conjugation action of on . By Sylow II, we know all -Sylow subgroups are conjugate, so the action is transitive, hence by orbit-stabiliser theorem is bijective to for all . Note that the stabliser is exactly the normaliser. So and therefore divides .
Now, restrict the action from to . We claim that the action of on has exactly one fixed point, namely . Firstly, since divides , . So the fixed point exists because of the lemma. Secondly, we show that is the only fixed point. Suppose is some -Sylow subgroup that is fixed by , that is, for all . Then . Thus, in , we have two -Sylow subgroups, and . By Sylow II, they must be conjugate in , but is normal in , so . Therefore, is the only fixed point of the action of on . By class equation, we have for some integer .
e.g. There is exactly one -Sylow subgroup in a group of order and it is normal. More generally, If with primes , then the -Sylow subgroup in is normal because the only integer of the form that divides is .
Examples of Sylow’s Theorems
e.g. We can use Sylow’s theorems to show that any group of order is isomorphic to . Note that , so by Sylow III, there is a unique -Sylow subgroup . Similarly, there is a unique -Sylow subgroup . Note that nontrivial elements of have order , while elements of have order , so they intersect only at the identity, i.e., . Therefore, by the proposition.