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nfain3

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Registered: 2010-10-02

Posts: 1

abstract algebra

can u help me with this Q?
prove that a a group of order 595 has a normal sylow 17-subgroup.

Alg Num Theory

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Registered: 2017-11-24

Posts: 693

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Re: abstract algebra

Let G be a group of order 595 = 5 × 7 × 17.

The number n of Sylow 17-subgroups divides 5 × 7 = 35 and is ≡ 1 (mod 17). Possibilities are 1 and 35.

Suppose 35. Then the intersection of all these subgroups (any two of which only have the identity in common since 17 is prime) would contain 1 + 35×(17−1) = 561 elements, i.e. 560 elements of order 17. Now, if P is a Sylow 17-subgroup and Q is a Sylow 5-subgroup, then PQ would be a subgroup of order |P||Q|/|P∩Q| = 85; moreover it would be cyclic, since P and Q are cyclic and their orders are coprime. But a cyclic subgroup of order 85 has φ(85) = 64 generators, i.e. 64 elements of order 85 – and we already have 560 elements of order 17.

Hence there can only be 1 Sylow 17-subgroup.

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