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120HW7 - HW 7 Solutions Chapter 4.5 Exercise 29 If G is a...

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Unformatted text preview: HW 7 Solutions May 24, 2009 Chapter 4.5 Exercise 29: If G is a non-abelian simple group of order < 100, prove that G u A 5 . Solution: OK, I’ll be honest here. I don’t particularly enjoy writing up these solutions, and this problem especially pisses me off. Forgive me if I take as many shortcuts as possible. Hopefully you’ve covered these cases either in class or you’ve read about them in your book. Groups of order p α are abelian or not simple. Groups of order 12 and order 30 are not sim- ple. If p < q < r are primes, then groups of order pq , p 2 q , pq 2 and pqr are not simple. Oh, and the book gives proof that any group of order 60 that is simple is isomorphic to A 5 , so we just have to show that groups of order 24 , 36 , 40 , 48 , 56 , 72 , 80 , 84 , 88 , 90 , 96 , and 100 are abelian or not simple. If I’m not mistaken, it turns out that all groups of order p α q β are abelian or not simple (Burnside’s Theorem). If we assume this, we only have groups of order 84 and 90 to rule out...but this would be akin to cheating. It is good trivia to know anyway. Some of these orders are quick and easy: 40, 84, 88, and 100. For in- stance, 40 equals 2 3 · 5. The number of Sylow 5-subgroups is congruent to 1 modulo 5 and divides 8. The only number that fits the bill is 1. So any group of order 40 has a normal subgroup isomorphic to Z / 5. An analogous argument shows that any group of order 88 must have a normal subgroup isomorphic to Z / 11 and any group of order 100 must have a normal sub- group or order 25. We get lucky on 84 as well, since its number of Sylow 7-subgroups is congruent to 1 modulo 7, but divides 12. Hence it has a normal subgroup isomorphic to Z / 7. Proposition 0.1. Let p and q be primes and suppose G is a group of order 1 p α q and the number of Sylow q-subgroups is p α , then G must have exactly one Sylow p-subgroup (which must be normal by the Sylow Theorem). Proof. The intersection of any two Sylow q-sugroups is the identity sub- group, so there must be p α ( q- 1) elements in G of order q . This leaves p α elements whose order is not equal to q . A Sylow p-subgroup has exactly that many elements, so evidently there can be at most one Sylow p-subgroup, in which case we’re done. This rules out groups of order 80 = 2 4 · 5 and 56 = 2 3 · 7. In the case of 80 for instance, the number of Sylow 5-subgroups is congruent to 1 mod5 and divides 16. Either there is one Sylow 5-subgroup, which must be normal, or there are 16 of them, in which case we use the proposition to conclude there is one normal Sylow 2-subgroup. Groups of order 56 are dealt with analogously. Now we’re left to rule out groups of orders 24 , 36 , 48 , 72 , 90, and 96. We’ll need another little proposition....
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120HW7 - HW 7 Solutions Chapter 4.5 Exercise 29 If G is a...

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