Page 138 - DMTH403_ABSTRACT_ALGEBRA

Unit 12: Sylow’s Theorems

Notes
Example: Show that there is no simple group of order 200.
Solution: Since 200 = 2 3 . 5 , the number of Sylow 5-subgroups is congruent to 1 mod 5 and a
2
divisor of 8. Thus there is only one Sylow 5-subgroup, and it is a proper nontrivial normal
subgroup.

Example: Show that a group of order 108 has a normal subgroup of order 9 or 27.
Solution: Let S be a Sylow 3-subgroup of G. Then [G:S] = 4, since |G| = 2 3 , so we can let G act
3
2
by multiplication on the cosets of S. This defines a homomorphism µ : G -> S , so it follows that
4
| µ(G) | is a divisor of 12, since it must be a common divisor of 108 and 24. Thus | ker(µ) |  9,
and it follows that ker(µ)  S, so | ker(µ) | must be a divisor of 27. It follows that | ker(µ) | = 9
or | ker(µ) | = 27.

Example: If p is a prime number, find all Sylow p-subgroups of the symmetric group S .
p
Solution: Since |S | = p!, and p is a prime number, the highest power of p that divides |S | is p.
p
p
Therefore, the Sylow p-subgroups are precisely the cyclic subgroups of order p, each generated
by a p-cycle. There are (p-1)! = p! / p ways to construct a p-cycle (a , . . . , a ). The subgroup
1
p
generated by a given p-cycle will contain the identity and the p-1 powers of the cycle. Two
different such subgroups intersect in the identity, since they are of prime order, so the total
number of subgroups of order p in S is (p-2)! = (p-1)! / (p-1).
p
Example: Prove that if G is a group of order 56, then G has a normal Sylow 2-subgroup
or a normal Sylow 7-subgroup.
Solution: The number of Sylow 7-subgroups is either 1 or 8. Eight Sylow 7-subgroups would
yield 48 elements of order 7, and so the remaining 8 elements would constitute the (unique)
Sylow 2-subgroup.

Example: Prove that if N is a normal subgroup of G that contains a Sylow p-subgroup of
G, then the number of Sylow p-subgroups of N is the same as that of G.

Solution: Suppose that N contains the Sylow p-subgroup P. Then since N is normal it also
contains all of the conjugates of P. But this means that N contains all of the Sylow p-subgroups
of G, since they are all conjugate. We conclude that N and G have the same number of Sylow
p-subgroups.

Example: Prove that if G is a group of order 105, then G has a normal Sylow 5-subgroup
and a normal Sylow 7-subgroup.

Solution: The notation n (G) will be used for the number of Sylow p-subgroups of G. Since 105
p
= 3 · 5 · 7, we have n (G) = 1 or 7, n (G) = 1 or 21, and n (G) = 1 or 15 for the numbers of Sylow
7
5
3
subgroups. Let P be a Sylow 5-subgroup and let Q be a Sylow 7-subgroup. At least one of these
subgroups must be normal, since otherwise we would have 21 · 4 elements of order 5 and 15 · 6
elements of order 7. Therefore, PQ is a subgroup, and it must be normal since its index is the
smallest prime divisor of |G|, so we can apply the result in the previous problem. Since PQ is
normal and contains a Sylow 5-subgroup, we can reduce to the number 35 when considering the
number of Sylow 5-subgroups, and thus n (G) = n (PQ) = 1. Similarly, since PQ is normal and
5
5
contains a Sylow 7-subgroup, we have n (G) = n (PQ) = 1.
7
7

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