Page 137 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes The more precise fact established in our proof of Sylow’s Second Theorem yields the following
useful result:
Corollary: If G is a finite group and p is a prime number, then any subgroup of G of p-power
order is contained in some Sylow p-subgroup.

Since G acts transitively by conjugation on Syl (G), and the stabilizer of P  Syl (G) is N (P), we
p
p
G
deduce that np = [G : N (P)] for any P  Syl (G).
G
p
Therefore:
Corollary: If G is a finite group and p is a prime number, let n be the number of Sylow
p
p-subgroups of G. Then the following are equivalent:
1. np = 1.
2. Every Sylow p-subgroup of G is normal.
3. Some Sylow p-subgroup of G is normal.

Example: By direct computation, find the number of Sylow 3-subgroups and the number
of Sylow 5-subgroups of the symmetric group S . Check that your calculations are consistent
5
with the Sylow theorems.
Solution: In S there are ( 5 · 4 · 3 ) / 3 = 20 three cycles. These will split up into 10 subgroups of
5
order 3. This number is congruent to 1 mod 3, and is a divisor of 5 · 4 · 2.
There are ( 5! ) / 5 = 24 five cycles. These will split up into 6 subgroups of order 5. This number
is congruent to 1 mod 5, and is a divisor of 4 · 3 · 2.

Example: How many elements of order 7 are there in a simple group of order 168?
3 7. The number of Sylow 7-subgroups must be congruent to 1 mod 7
Solution: First, 168 = 2 3 . .
and must be a divisor of 24. The only possibilities are 1 and 8. If there is no proper normal
subgroup, then the number must be 8. The subgroups all have the identity in common, leaving
8 · 6 = 48 elements of order 7.

Example: Prove that a group of order 48 must have a normal subgroup of order 8 or 16.
Solution: The number of Sylow 2-subgroups is 1 or 3. In the first case there is a normal subgroup
of order 16. In the second case, let G act by conjugation on the Sylow 2-subgroups. This produces
a homomorphism from G into S . Because of the action, the image cannot consist of just 2
3
elements. On the other hand, since no Sylow 2-subgroup is normal, the kernel cannot have 16
elements. The only possibility is that the homomorphism maps G onto S , and so the kernel is a
3
normal subgroup of order 48 / 6 = 8.

Example: Let G be a group of order 340. Prove that G has a normal cyclic subgroup of
order 85 and an abelian subgroup of order 4.
2 . .
Solution: First, 340 = 2 5 17. There exists a Sylow 2-subgroup of order 4, and it must be abelian.
No divisor of 68 = 2 2 . 17 is congruent to 1 mod 5, so the Sylow 5-subgroup is normal. Similarly,
then Sylow 17-subgroup is normal. These subgroups have trivial intersection, so their product
.
is a direct product, and hence must be cyclic of order 85 = 5 17. The product of two normal
subgroups is again normal, so this produces the required normal subgroup of order 85.

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