sep07 - of 2 so if there was a second Sylow 2-subgroup we...

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Math 5125 Wednesday, September 7 September 7, Ungraded Homework Exercise 4.5.13 on page 147 Prove that a group of order 56 has a normal Sylow p - subgroup for some prime p dividing its order. Let G be a group of order 56 and suppose it does not have a normal Sylow 7-subgroup. We have 56 = 2 3 · 7. Therefore the number of Sylow 7-subgroups is congruent to 1 mod 7 and divides 8. Since G does not have a normal Sylow 7-subgroup, 1 is ruled out, consequently it has exactly 8 Sylow 7-subgroups. We now count the number of elements of order 7. Since there are 6 elements of order 7 in a Sylow 7-subgroup and any two Sylow 7-subgroups intersect in 1 by Lagrange, there are 8 · 6 elements of order 7 in G . There are 56 - 48 = 8 elements remaining. If G has at least two Sylow 2-subgroups, then it would have at least 9 elements whose order is a power of 2 (a Sylow 2-subgroup has 8 elements of order a power
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Unformatted text preview: of 2, so if there was a second Sylow 2-subgroup, we would get at least one more element whose order was a power of 2). Thus G has only one Sylow 2-subgroup 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 11-subgroup and let N denote its normalizer. Then the number of Sylow 11-subgroups is congruent to 1 mod 11 and divides 12, so is 1 or 12. But if there is exactly one Sylow 11-subgroup, then it is normal which contradicts the hypothesis that G is simple. Therefore G has 12 Sylow 11-subgroups; 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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This note was uploaded on 01/02/2012 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.

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