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Unformatted text preview: Solutions to (Selected) HW Problems Math 551 – Fall 2007 December 5, 2007 Please read: I will try to post here a few solutions. The new solutions will be added to this same file. Disclaimer: I will have to put these solutions together rather quickly, so they are subject to typos and conceptual mistakes. (I expect you to be a lot more careful when doing your HW than I when preparing these.) You can contact me if you think that there is something wrong and I will fix the file if you are correct. Homework 3 Section 3.2 10. Note: HK might not be a subgroup of G . We have  G : H ∩ K  =  G : H  H : H ∩ K  = m  H : H ∩ K  (1) =  G : K  K : H ∩ K  = n  K : H ∩ K  (2) So, if  G : H ∩ K  < ∞ , then it is a common multiple of m,n . So, in any case: lcm( m,n ) ≤  G : H ∩ K  . Claim:  H : H ∩ K  ≤  G : K  . Proof. Let hH ∩ K and h H ∩ K be distinct cosets of H ∩ K in H . If hK = h K , then h = h k for some k ∈ K . Thus ( h ) 1 h = k ∈ H ∩ K and hH ∩ K = h H ∩ K . This implies that  H : H ∩ K  ≤  G : K  = n . 1 The claim together with (1), imply that  G : H ∩ K  = m  H : H ∩ K  ≤ mn . 14. Suppose that H / S 4 and  H  = 8. If H has a 4cycle, then it has all 4cycles. But there are 4 · 3 · 2 · 1 4 = 6 of those. With the identity, we’d have seven elements. Then, there must be another element in H , which implies that all elements are of that structure. In fact, we have: Cycle Structure Number of Elements 1 1 ( ab ) 4 · 3 2 = 6 ( ab )( cd ) 1 2 ( 4 · 3 2 · 1 ) = 3 ( abc ) 4 · 3 · 2 3 = 8 ( abcd ) 4 · 3 · 2 · 1 4 = 6 Thus, we cannot have another element, or we would have all the elements of the same cycle structure, yielding too many elements. For  H  = 3, we must have a 3cycle in H [by Lagrange’s Theorem], and since H / G , we would have to have all 8 3cycles in H , which are again too many. 18. Let π : G → G/N , so we can restrict this map to H , i.e., consider π  H : H → H/N , which gives us that  im( π  H )    G/N  . Also by the “Fist Isomorphism Theorem”, we have  im( π  H )    H  , which implies that  im( π  H )  = 1. Thus H ⊆ ker( π ) = N . 19. Suppose  H  =  N  . Then (  H  ,  G : N  ) = (  N  ,  G : N  ) = 1, and by the previous problem, we get H ⊆ N . Since  H  =  N  < ∞ , H = N . 21. Suppose  Q : H  = n < ∞ . Then for all p/q ∈ Q , we have that n ( p/q ) ∈ H . Thus, for any p/q ∈ Q , n p q · n = p q ∈ H, and so H = Q . If  Q / Z : ¯ H  < n < ∞ , then there exists H ⊆ Q , Z ⊆ H such that Q /H ∼ = ( ¯ Q / Z ) / ¯ H . By the above, this implies that H = Q , and so ¯ H = Q / Z . 2 Section 3.3 1. Let det : GL n ( F ) → F × . Then det is a homomorphism, it’s onto, since we can multiply a row of the identity by any element of k ∈ F × , giving a matrix of determinant k . Hence, ker det = SL n ( F ), and by the First Isomorphism Theorem , we have that  GL n ( F ) / SL n ( F )...
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This note was uploaded on 11/19/2010 for the course ANTH 122 taught by Professor 323 during the Spring '10 term at Centennial College.
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