Unformatted text preview: of 2, so if there was a second Sylow 2subgroup, we would get at least one more element whose order was a power of 2). Thus G has only one Sylow 2subgroup and so it must be normal. This completes the proof. Prove that a group of order 1452 cannot be simple. Suppose G is a simple group of order 1452 = 2 2 · 3 · 11 2 . Let P be a Sylow 11subgroup and let N denote its normalizer. Then the number of Sylow 11subgroups is congruent to 1 mod 11 and divides 12, so is 1 or 12. But if there is exactly one Sylow 11subgroup, then it is normal which contradicts the hypothesis that G is simple. Therefore G has 12 Sylow 11subgroups; this means that  G : N  = 12. Since G is simple, N 6 = G and  G  6 = 2, we deduce that G is isomorphic to a subgroup of A 12 . This is a contradiction because 11 2 ± ±  G  and 11 2 A 12  ....
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 Fall '07
 PALinnell
 Math, Algebra, Group Theory, Cyclic group, Sylow, Sylow 11subgroups

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