Sylow subgroups exist
Contents
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
Other parts of Sylow's theorem
- Sylow implies order-dominating: Given any
-Sylow subgroup and any
-subgroup, the
-Sylow subgroup contains a conjugate of the
-subgroup.
- Sylow implies order-conjugate: All
-Sylow subgroups are conjugate. This follows directly from the previous part.
- Congruence condition on Sylow numbers: The number of
-Sylow subgroups is congruent to
modulo
.
Stronger forms of existence
- Every Sylow subgroup intersects the center nontrivially or is contained in a centralizer: Every Sylow subgroup of a group either has a nontrivial intersection with the center of the group, or it is contained in the centralizer of some non-central element.
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 group: 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
.
Analogues in other algebraic structures
- Sylow subloops exist for Sylow primes in finite Moufang loop
- 2-Sylow subloops exist in finite Moufang loop
- 3-Sylow subloops exist in finite Moufang loop
- Hall subloops exist in finite solvable Moufang loop
Facts used
- Lucas' theorem (more specifically, Lucas' theorem prime power case)
- 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.
- Lagrange's theorem
- Class equation of a group
- Cauchy's theorem for abelian groups
- Central implies normal: Any subgroup inside the center is normal.
Proof
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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 ![]() ![]() ![]() |
|||
2 | The size of ![]() ![]() |
Fact (1) | The size of ![]() ![]() ![]() | |
3 | ![]() ![]() |
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 ![]() ![]() ![]() ![]() | ||
4 | The action of ![]() ![]() ![]() ![]() |
Step (2) | The size of ![]() ![]() ![]() ![]() | |
5 | Let ![]() ![]() ![]() ![]() ![]() |
|||
6 | The index of ![]() ![]() ![]() |
Fact (2) | Steps (4), (5) | Follows directly from fact (2). |
7 | The index of ![]() ![]() ![]() |
Steps (4), (6) | By step (4), the size of ![]() ![]() ![]() ![]() ![]() ![]() | |
8 | The index of ![]() ![]() ![]() |
Fact (3) | Step (6) | By fact (3) (Lagrange's theorem), the index of ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
9 | The index of ![]() ![]() ![]() |
Steps (1), (4), (5) | The union of ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() | |
10 | The index of ![]() ![]() ![]() ![]() ![]() ![]() |
Steps (8), (9) | The index of ![]() ![]() ![]() ![]() ![]() |
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 ![]() ![]() |
Facts (5), (6) | Since ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() | |
2 | ![]() ![]() ![]() ![]() |
Step (1) | Since ![]() ![]() ![]() ![]() ![]() ![]() ![]() | |
3 | Let ![]() ![]() ![]() ![]() |
Steps (1), (2) | Indeed, the order of ![]() ![]() ![]() ![]() |
Case that does not divide the order of
Step no. | Construction/assertion | Facts used | Previous steps used | Explanation |
---|---|---|---|---|
1 | There exists ![]() ![]() ![]() ![]() ![]() |
Since ![]() ![]() ![]() ![]() ![]() ![]() | ||
2 | ![]() ![]() ![]() ![]() ![]() |
Fact (3) | Step (1) | Since ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() |
3 | ![]() ![]() |
Step (2) | This follows by the induction hypothesis. | |
4 | ![]() ![]() ![]() |
Step (3) | A subgroup of order ![]() ![]() ![]() ![]() |