# sols - Solutions to (Selected) HW Problems Math 551 –...

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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 4-cycle, then it has all 4-cycles. 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 3-cycle in H [by Lagrange’s Theorem], and since H / G , we would have to have all 8 3-cycles 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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sols - Solutions to (Selected) HW Problems Math 551 –...

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