## § 7.4 The Sylow Theorems: solved problems

13. By direct computation, find the number of Sylow 3-subgroups and the number of Sylow 5-subgroups of the symmetric group S5. Check that your calculations are consistent with the Sylow theorems.

Solution: In S5 there are ( 5 · 4 · 3 ) / 3 = 20 three cycles. These will split up into 10 subgroups of 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.

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14. How many elements of order 7 are there in a simple group of order 168?

Solution: First, 168 = 23 · 3 · 7. The number of Sylow 7-subgroups must be congruent to 1 mod 7 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.

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15. 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 S3. Because of the action, the image cannot consist of just 2 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 S3, and so the kernel is a normal subgroup of order 48 / 6 = 8.

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16. 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.

Solution: First, 340 = 22 · 5 · 17. There exists a Sylow 2-subgroup of order 4, and it must be abelian. No divisor of 68 = 22 · 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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17. Show that there is no simple group of order 200.

Solution: Since 200 = 23 · 52, the number of Sylow 5-subgroups is congruent to 1 mod 5 and a divisor of 8. Thus there is only one Sylow 5-subgroup, and it is a proper nontrivial normal subgroup.

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18. 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| = 22 33, so we can let G act by multiplication on the cosets of S. This defines a homomorphism µ : G -> S4, so it follows that | 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 from Exercise 7.3.1 of the text that ker(µ) S, so | ker(µ) | must be a divisor of 27. It follows that | ker(µ) | = 9 or | ker(µ) | = 27.

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19. If p is a prime number, find all Sylow p-subgroups of the symmetric group Sp.

Solution: Since |Sp| = p!, and p is a prime number, the highest power of p that divides |Sp| is 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 (a1, . . . , ap). The subgroup 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 indentity, since they are of prime order, so the total number of subgroups of order p in Sp is (p-2)! = (p-1)! / (p-1).

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20. 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.

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21. 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.

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22. 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 np(G) will be used for the nubmer of Sylow p-subgroups of G. Since 105 = 3 · 5 · 7, we have n3(G) = 1 or 7, n5(G) = 1 or 21, and n7(G) = 1 or 15 for the numbers of Sylow 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 n5(G) = n5(PQ) = 1. Similarly, since PQ is normal and contains a Sylow 7-subgroup, we have n7(G) = n7(PQ) = 1.

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