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**nfain3****Member**- Registered: 2010-10-02
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can u help me with this Q?

prove that a a group of order 595 has a normal sylow 17-subgroup.

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**Alg Num Theory****Member**- Registered: 2017-11-24
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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.

Me, or the ugly man, whatever (3,3,6)

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