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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 )...
View Full Document

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.

Page1 / 76

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online