HW10_Solutions

HW10_Solutions - Math 541 Solutions to HW#10 1 Prove that...

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Math 541 Solutions to HW #10 1. Prove that if G is a simple group such that 61 ≤| G |≤ 70, then G is a cyclic group. [Feel free to use the Sylow theorems.] Referencing the lemmas from homework 9, we can create the following table: Order Simple/Not Simple Reason 61 Simple Lemma P 62 Not Simple Lemma PQ: 62 = 2 * 31 63 Not Simple Sylow’s Third Theorem: n 7 | 63 implies n 7 = 1, 3, 7, 9, 21, or 63. Of these, only 1 1 (mod 7). Thus, by the corollary to Sylow’s Third Theorem, there is a normal subroup of order 7. 64 Not Simple Lemma Z: 64 = 2 6 65 Not Simple Lemma PQ: 65 = 5 * 13 66 Not Simple Sylow’s Third Theorem: n 11 | 66 implies n 11 = 1, 2, 3, 6, 11, 22, 33, or 66. Of these, only 1 1 (mod 1)1. Thus, by the corollary to Sylow’s Third Theorem, there is a normal subgroup of order 11. 67 Simple Lemma P 68 Not Simple Sylow’s Third Theorem: n 17 | 68 implies n 17 = 1, 2, 4, 17, 34, or 68. Of these, only 1 1 (mod 17). Thus, by the corollary to Sylow’s Third Theorem, there is a normal subgroup of order 17. 69 Not Simple Lemma PQ: 69 = 3 * 23 70 Not Simple Sylow’s Third Theorem: n 7 | 70 implies n 7 = 1, 2, 5, 7, 10, 14, 35, or 70. Of these, only 1 1 (mod 7). Thus, by the corollary to Sylow’s Third Theorem, there is a normal subgroup of order 7. Notice that the only simple groups are those that have prime order. Since every group of prime order is cyclic, the proof is complete. 2. Let A and B be normal subgroups of a group G such that A B = { e } . Prove that ab = ba for all a A and b B . [Hint: Prove that aba - 1 b - 1 A B .] Note first that for all a A and b B , we have aba - 1 B and ba - 1 b - 1 A . Correspondingly, ( aba - 1 ) b - 1 Bb - 1 = B and a ( ba - 1 b - 1 ) aA = A . Thus, aba - 1 b - 1 A B . Since A B = { e } , we know that aba - 1 b - 1 = e , and therefore that ab = ba . This completes the proof. 3. Let G be a group of size 15. Let a be an element of order 3 and b be an element of order 5. Set A = h a i and B = h b i . (a) Prove that AB = G . Since A and B have prime orders 3 and 5, each non-identity element of A and B also has an order of 3 and 5, respectively. Therefore A B = { e } , for otherwise we could conclude | A | = 5 or | B | = 3, both of which would be contradictions. One way to accomplish our task is to determine if the set AB = { ab | a A and b B } has size 15. We can do this using properties of cosets. Let A = { e, a 1 , a 2 } . Then B = eB 6 = a 1 B and B = eB 6 = a 2 B , since a 1 , a 2 / B . Since different cosets are disjoint, the only case in 1
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which | AB | < 15 is if a 1 B = a 2 B . Seeking a contradiction, suppose a 1 B = a 2 B . Then for every b 1 B , there exists a b 2 B such that a 1 b 1 = a 2 b 2 with b 1 6 = b 2 (since otherwise we get a 1 = a 2 ), and consequently a 1 b 1 b - 1 2 = a 2 A . Let b
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