# Sylow subgroups exist

## Statement

Let  be a Finite group (?) and  be a prime number. Then, there exists a -Sylow subgroup (?)  of : a subgroup whose order is a power of  and whose index is relatively prime to .

Note that when  does not divide the order of , the -Sylow subgroup is trivial, so the statement gives interesting information only when  divides the order of . This statement is often viewed as a part of a more general statement called Sylow's theorem.

## Related facts

### Analogues for Hall subgroups

The analogous statement does not hold for Hall subgroups. A Hall subgroup is a subgroup whose order and index are relatively prime. A -Hall subgroup is a Hall subgroup such that the set of primes dividing its order is contained in , and the set of primes dividing its index is disjoint from .

• Hall subgroups need not exist: Given a set  of primes, there may not exist a -Hall subgroup.
• Hall subgroups exist in finite solvable: If, however, the finite group is solvable, then it has -Hall subgroups for all prime sets .
• Hall's theorem: This states that a finite group is solvable if and only if it has -Hall subgroups for every prime set .

## Facts used

1. Lucas' theorem (more specifically, Lucas' theorem prime power case)
2. Fundamental theorem of group actions: There is a bijection between the coset space of the stabilizer of an element and the orbit of that element. In particular, the size of the orbit equals the index of the stabilizer.
3. Lagrange's theorem
4. Class equation of a group
5. Cauchy's theorem for abelian groups
6. Central implies normal: Any subgroup inside the center is normal.

## Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

### Proof by action on subsets

Given: A finite group  of order  where  is prime,  is a nonnegative integer, and  does not divide .

To prove:  has a subgroup of order .

Proof: Here are some observations regarding this action:

Step no. Construction/assertion Facts used Previous steps used Explanation
1 Let  be the set of all subsets of  of size .
2 The size of  is relatively prime to  Fact (1) The size of  is the binomial coefficient , and an appeal to Lucas's theorem (fact (1)) reveals that its value is relatively prime to .
3  acts on  by left multiplication. First, note that a group acts on the set of all its subsets by left multiplication. Further, since the left multiplication maps are bijective, they preserve the sizes of subsets. In particular, any subset of size  gets mapped to a subset of size . Thus, we can restrict the action to the set  of subsets of size .
4 The action of  on  has at least one orbit, say , whose size is relatively prime to  Step (2) The size of  equals the sum of sizes of the orbits under the -action. Since the total is relatively prime to , at least one of the numbers has to be relatively prime to .
5 Let  be a member of such an orbit  and  be the stabilizer of  in .
6 The index of  in  equals the size of  Fact (2) Steps (4), (5) Follows directly from fact (2).
7 The index of  in  is relatively prime to . Steps (4), (6) By step (4), the size of  is relatively prime to . By step (6), this equals the index of . Thus, the index of  in  is relatively prime to .
8 The index of  in  divides . Fact (3) Step (6) By fact (3) (Lagrange's theorem), the index of  in  divides the order of , which is . By step (6), the index of  in  is relatively prime to . Hence, it must divide .
9 The index of  in  is at least . Steps (1), (4), (5) The union of , is the whole group . Since each  has size , there are at least  translates of  in . Thus,  has size at least .
10 The index of  in  is exactly , so  is a -Sylow subgroup of . Steps (8), (9) The index of  in  divides  and is at least . The only way both of these could be true is if it equals .

Interestingly, the only nontrivial result we use here (Lucas' theorem) can itself be proved using group theory by doing the above proof in reverse taking a cyclic group (although a purely algebraic proof also exists). ,

### Proof by conjugation action

Given: A finite group  of order , where  is prime,  is a nonnegative integer, and  does not divide .

To prove:  has a subgroup of order .

Proof: We prove the claim by induction on the order of . Specifically, we assume that the result is true for all groups of order smaller than the order of .

For the case , the proof is direct since the trivial group is a subgroup of order . Thus, we assume . In particular,  divides the order of .

Consider the class equation of  (fact (4)):



where  are the conjugacy classes of non-central elements and  is an element of  for each .

We consider two cases:

Case that  divides the order of  :

Step no. Construction/assertion Facts used Previous steps used Explanation
1 There exists a normal subgroup of order  in  Facts (5), (6) Since  is Abelian, fact (5) yields that it has a subgroup  of order . Since  is in the center,  is normal in  (by fact (6)). Thus,  is a normal subgroup of  of order .
2  has a -Sylow subgroup, in particular, a subgroup  of order  Step (1) Since  has order ,  has order . This is strictly smaller than the order of , so induction applies and we get a -Sylow subgroup, whose order is .
3 Let  be the quotient map. Then  is a -Sylow subgroup of  Steps (1), (2) Indeed, the order of  is the product of the order of  and the order of , which is therefore .

Case that  does not divide the order of 

Step no. Construction/assertion Facts used Previous steps used Explanation
1 There exists  such that  does not divide the index  of  in  Since  divides the order of ,  cannot divide the index of every , otherwise the class equation would yield that  divides the order of .
2  is a proper subgroup of  whose order is a multiple of , so  is the largest power of  dividing its order. Fact (3) Step (1) Since  is non-central,  is proper in . Further, since  is relatively prime to , Lagrange's theorem (fact (3)) yields that the order of  is , which is a multiple of .
3  contains a subgroup of order  Step (2) This follows by the induction hypothesis.
4  contains a subgroup of order , and hence a -Sylow subgroup. Step (3) A subgroup of order  in  is also a subgroup of order  in .