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Homework8

# Homework8 - Homework 8 Solutions 1(a False e.g S3 has 9...

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Homework 8 Solutions 1. (a) False. e.g. S 3 has 9 Sylow 2-subgroups. (b) True. G abelian implies all subgroups are normal, which implies that each Sylow sub- group is unique. (c) True. Sylow’s Theorem part 1. (d) True. If y = gx then G y = gG x g - 1 . (e) False. e.g. G = S 3 and H = (12) . N G ( H ) = H G . (f) True. Definition of normalizer. (g) False. If the Sylow p -subgroup isn’t normal then this doesn’t hold. (h) True. The size of the Sylow p -subgroup is p n where n is the largest power of p appearing in the order of the group. Since | G | = | H | , n is the same for both of them. (i) False. 8 1 (mod 5). 2. Let G = S 3 × S 3 . Subgroups of size 4 are the Sylow 2-subgroups. Let P = (12) × (12) , Q = (13) × (13) and R = (12) × (13) . P , Q and R are all Sylow 2-subgroups. P Q = { ((1) , (1)) } = { e } but P R = { e, ((12) , (1)) } . 3. (a) 300 = 2 2 · 3 · 5 2 . n 5 1 (mod 5) and n 5 | 12. Thus n 5 = 1 or 6. If n 5 = 1 then the Sylow 5-subgroup is normal and thus the group isn’t simple. So we need to deal with the case of n 5 = 6 Suppose n 5 = 6. This implies [ G : N ( P )] = 6 where P is one of the Sylow 5-subgroups (it doesn’t matter which one). Generalized Cayley’s theorem implies that there exists

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Homework8 - Homework 8 Solutions 1(a False e.g S3 has 9...

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